Bài 1: Tìm x thuộc Q, biết 1, $\frac{x 1}{2} \frac{x 1}{3} \frac{x 1}{4}=\frac{x 1}{5} \frac{x 1}{6}$ 2,\(\frac{x 1}{2009} \frac{x 2}{2008} \frac{x 3}{2007}=\frac{x 10}{2000} \frac{x 11}{1999} \frac...

2

NV

Nguyễn Võ Anh Nguyên

11 tháng 8 2017

1/$\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}$

$\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0$

$\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0$

Vì$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0$nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10

DV

Đỗ Viết Lâm Duy

VIP

25 tháng 6 2023

1/ $2 x + 1 + 3 x + 1 + 4 x + 1 = 5 x + 1 + 6 x + 1$

$\Leftrightarrow 2 x + 1 + 3 x + 1 + 4 x + 1 - 5 x + 1 - 6 x + 1 = 0$

$\Leftrightarrow (x + 1) (2 1 + 3 1 + 4 1 - 5 1 - 6 1) = 0$

Vì $2 1 + 3 1 + 4 1 - 5 1 - 6 1 > 0$ nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10

1/ $2 x + 1 + 3 x + 1 + 4 x + 1 = 5 x + 1 + 6 x + 1$

$\Leftrightarrow 2 x + 1 + 3 x + 1 + 4 x + 1 - 5 x + 1 - 6 x + 1 = 0$

$\Leftrightarrow (x + 1) (2 1 + 3 1 + 4 1 - 5 1 - 6 1) = 0$

Vì $2 1 + 3 1 + 4 1 - 5 1 - 6 1 > 0$ nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10

DP

Đoàn Phương Linh

12 tháng 8 2018

Tìm x biết:

1) $\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}$

2) $\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

2

KT

Không Tên

12 tháng 8 2018

1) $\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}$

<=> $\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0$

<=> $x+1=0$ (do 1/2 + 1/3 + 1/4 - 1/5 - 1/6 khác 0)

<=> $x=-1$

Vậy...

KT

Không Tên

12 tháng 8 2018

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

<=> $\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1$

<=> $\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}$

<=> $\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

<=> $x+2010=0$ (do 1/2009 + 1/2008 + 1/2007 - 1/2000 - 1/1999 - 1/1998 khác 0)

<=> $x=-2010$

Vậy....

NT

Nguyen Thi Thanh Thao

28 tháng 9 2016

Tìm x biết :

a) $-\frac{2}{3}.x+4=-12$

b) $-\frac{3}{4}+\frac{1}{4}:x=-3$

c) $\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}$

d)$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

1

ND

Nguyễn Đình Dũng

28 tháng 9 2016

d) $\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

<=> $\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0$

<=> $\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)$

<=> $\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0$

<=>$\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

<=> x+2010 = 0 vì $\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0$

<=> x = -2010

ND

Nguyễn Đình Dũng

28 tháng 9 2016

Làm câu khó nhất rồi, còn lại tự làm nha <(") /_\

A

ANYWAY

10 tháng 7 2019

Tìm x

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

1

NH

Nhật Hạ

10 tháng 7 2019

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

$\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0$

$\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

$\Leftrightarrow x+2010=0$

$\Leftrightarrow x=-2010$

S

Sagittarus

16 tháng 8 2015

tìm x biết:

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

3

TV

Tiến Vỹ

2 tháng 9 2017

x=-2010 nhé

chúc bạn hc tốt

DL

Dương Lam Hàng

7 tháng 6 2018

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

$\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+x+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1$

$\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}$

$\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}$

$\Rightarrow\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

Vì $\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0$

Nên x + 2010 = 0 => x = -2010

PQ

Phí Quỳnh Anh

13 tháng 7 2016

a)$\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+$$\frac{x+1}{6}$

b)$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=$$\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

c)$2.2^2.2^3.2^4.........2^x=1024$$\left(x\in z\right)$

Tìm x

3

DT

Đinh Thùy Linh

14 tháng 7 2016

b./ $\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0$

$\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$(b)

Mà $\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0$

(b) $\Leftrightarrow x+2010=0\Leftrightarrow x=-2010$

DT

Đinh Thùy Linh

14 tháng 7 2016

a./

$\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.$

$\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0$(a)

Mà $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0$

(a) $\Leftrightarrow x+1=0\Leftrightarrow x=-1$

G

Giang シ)

VIP

18 tháng 8 2021

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

Giúp mik với !! mik ko hiểu bài này !! giúp mik nhé

6

A

Anh2Kar六

18 tháng 8 2021

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.$

$\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.$(cộng 2 vế cho 3)

$\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.$

$\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.$

$\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0$

x+2010=0

x=-2010

MA

Mai Anh Nguyen

18 tháng 8 2021

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

$\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)$

$=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}$

$=0$

$\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

$\Leftrightarrow x+2010=0$

$\Leftrightarrow x=-2010$

SS

★ςɧάήɧ ςÀ κɧỊά★

22 tháng 9 2020

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

GIÚP MÌNH NHÉ ,MÌNH CẦN GẤP.

3

NM

Nguyễn Minh Đăng

22 tháng 9 2020

P/s: Công vào 6 phân thức trên, mỗi phân thức công thêm 1 rồi quy đồng lên ta được:

$\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}$

$\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

Ta xét: $\hept{\begin{cases}\frac{1}{2009}< \frac{1}{2000}\\\frac{1}{2008}< \frac{1}{1999}\\\frac{1}{2007}< \frac{1}{1998}\end{cases}}\Rightarrow\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0$

=> $x+2010=0\Rightarrow x=-2010$

Vậy x = -2010

KN

Khánh Ngọc

22 tháng 9 2020

$\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}$

$\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)$

$=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}$

$\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0$

$\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0$

$\Leftrightarrow x+2010=0$

$\Leftrightarrow x=-2010$

K

Kitty

20 tháng 7 2018

tìm...

Đọc tiếp

0

TM

Trần Minh Thắng

22 tháng 8 2020

...

Đọc tiếp

3

XO

Xyz OLM

22 tháng 8 2020

1) Ta có : $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

$\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0$

Vì $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0$

=> x + 1 = 0

=> x = - 1

b) $\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}$

=> $\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)$

=> $\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}$

=> $\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0$

Vì $\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0$

=> x + 2010 = 0

=> x = -2010

c) $\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}$

$\Rightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)+\left(\frac{x+1969}{69}-1\right)$

=> $\frac{x+1900}{45}+\frac{x+1900}{54}=\frac{x+1900}{75}+\frac{x+1900}{69}$

=> $\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0$

=> $x+1900=0\left(\text{Vì }\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\right)$

=> x = -1900

d) $\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}$

=> $\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)$

=> $\frac{x+2028}{10}+\frac{x+2028}{9}=\frac{x+2028}{8}+\frac{x+2028}{7}$

=> $\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0$

=> x + 2028 = 0 $\left(\text{Vì }\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\right)$

=> x = -2028

TD

Tạ Đức Hoàng Anh

22 tháng 8 2020

1) Ta có: $\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

$\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0$

$\Leftrightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0$

+ TH₁: $x+1=0$$\Leftrightarrow$$x=-1$

+ TH₂: $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0$

Vì $\hept{\begin{cases}\frac{1}{10}>\frac{1}{13}\\\frac{1}{11}>\frac{1}{14}\\\frac{1}{12}>0\end{cases}}$$\Rightarrow$$\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}$

$\Rightarrow$$\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$

mà $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}=0$

$\Rightarrow$Phương trình trên vô nghiệm

Vậy $x=-1$

2) Ta có: $\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}$

$\Leftrightarrow\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+2}{2008}+1\right)-\left(\frac{x+1}{2009}+1\right)=0$

$\Leftrightarrow\frac{x+2010}{2006}+\frac{x+2010}{2007}-\frac{x+2010}{2008}-\frac{x+2010}{2009}=0$

$\Leftrightarrow\left(x+2010\right).\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0$

+ TH₁: $x+2010=0$$\Leftrightarrow$$x=-2010$

+ TH₂: $\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0$

Vì $\hept{\begin{cases}\frac{1}{2006}>\frac{1}{2008}\\\frac{1}{2007}>\frac{1}{2009}\end{cases}}$$\Rightarrow$$\frac{1}{2006}+\frac{1}{2007}>\frac{1}{2008}+\frac{1}{2009}$

$\Rightarrow$$\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>0$

mà $\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}=0$

$\Rightarrow$Phương trình trên vô nghiệm

Vậy $x=-2010$

3) Ta có: $\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}$

$\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)-\left(\frac{x+1975}{75}-1\right)-\left(\frac{x+1969}{69}-1\right)=0$

$\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0$

$\Leftrightarrow\left(x+1900\right).\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0$

+ TH₁: $x+1900=0$$\Leftrightarrow$$x=-1900$

+ TH₂: $\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0$

Vì $\hept{\begin{cases}\frac{1}{45}>\frac{1}{75}\\\frac{1}{54}>\frac{1}{69}\end{cases}}$$\Rightarrow$$\frac{1}{45}+\frac{1}{54}>\frac{1}{75}+\frac{1}{69}$

$\Rightarrow$$\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}>0$

mà $\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}=0$

$\Rightarrow$Phương trình trên vô nghiệm

Vậy $x=-1900$

4) Ta có: $\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}$

$\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)-\left(\frac{x-95}{9}-1\right)-\left(\frac{x-93}{11}-1\right)=0$

$\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0$

$\Leftrightarrow\left(x-104\right).\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0$

+ TH₁: $x-104=0$$\Leftrightarrow$$x=104$

+ TH₂: $\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0$

Vì $\hept{\begin{cases}\frac{1}{5}>\frac{1}{7}\\\frac{1}{9}>\frac{1}{11}\end{cases}}$$\Rightarrow$$\frac{1}{5}+\frac{1}{7}>\frac{1}{9}+\frac{1}{11}$

$\Rightarrow$$\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}>0$

mà $\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}=0$

$\Rightarrow$Phương trình trên vô nghiệm

Vậy $x=104$

5) Ta có: $\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}$

$\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)-\left(\frac{x+2012}{8}+2\right)-\left(\frac{x+2014}{7}+2\right)=0$

$\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0$

$\Leftrightarrow\left(x+2028\right).\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0$

+ TH₁: $x+2028=0$$\Leftrightarrow$$x=-2028$

+ TH₂: $\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0$

Vì $\hept{\begin{cases}\frac{1}{10}< \frac{1}{8}\\\frac{1}{9}< \frac{1}{7}\end{cases}}$$\Rightarrow$$\frac{1}{10}+\frac{1}{9}< \frac{1}{8}+\frac{1}{7}$

$\Rightarrow$$\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}< 0$

mà $\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}=0$

$\Rightarrow$Phương trình trên vô nghiệm

Vậy $x=-2028$

Chúc bn hok tốt nha