giai các phương trình sau:a,$\frac{1-x}{2013}=1 \frac{2-x}{2012}-\frac{x}{2014}$b,$\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}$c,$\frac{x-a-b}{c} \frac{x-b-c}{a} \frac{x-a-c}{b}=3$d,(x 3)...

0

KT

Kiều Thị Hoa

21 tháng 4 2019

Giải Phương trình :

$a,\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}$

$b,\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}$

$c,\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-a-c}{b}=3$

0

NH

Nữ ❣√ Hoàng ♛⇝ Selfia ✔♫

21 tháng 4 2019

Giải pt :

$a,\frac{1-x}{2013}=1+\frac{2-x}{2014}-\frac{x}{2014},$

$b,\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}$

$c,\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-a-c}{b}=3$

1

GL

ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★

21 tháng 4 2019

$c,\frac{x-a-b}{c}-1+\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1=0.$

$\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}=0$

$\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0$

=>$\orbr{\begin{cases}a+b+c=x\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\end{cases}}$

Vậy.......

HL

Hiển Lê Sinh

10 tháng 5 2018

Giải phương trình sau:

a) $\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}$

b) $\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}$

3

NT

nguyễn thị lan hương

10 tháng 5 2018

a,$\Leftrightarrow\left(\frac{1-x}{2013}+1\right)=\left(\frac{2-x}{2012}+1\right)-\left(1-\frac{x}{2014}\right)$

$\Leftrightarrow\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{2014-x}{2014}$

$\Leftrightarrow\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}$=0

$\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\right)=0$

$\Leftrightarrow x=2014\left(do.cái.còn.lại.\ne0\right)$

b,tương tự +1 vào cái thứ nhất ,+1 vào cái thứ 2,1- vào cái thứ 3 được x=2013

CV

cao van duc

10 tháng 5 2018

ban oi them bot sai roi

TB

Trôi Bánh

9 tháng 8 2015

phần a : $\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}=\frac{x+2}{2013}+\frac{x+2}{2014}$

phần b :$\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}$

#Toán lớp 6

1

DN

Đỗ Ngọc Hải

9 tháng 8 2015

$a,\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}=\frac{x+2}{2013}+\frac{x+2}{2014}$

$\Leftrightarrow\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}-\frac{x+2}{2013}-\frac{x+2}{2014}=0$

$\Leftrightarrow\left(x+2\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0$

$\text{Mà }\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\text{ nên:}$

$\Leftrightarrow x+2=0$

$\Leftrightarrow x=-2$

$b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}$

$\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1$

$\Leftrightarrow \frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0$

$\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0$

$M\text{à}:\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0 n\text{ê}n:$

$x+2004=0$

$\Leftrightarrow x=-2004$

LT

Lãng Tử Buồn

6 tháng 7 2019

Giải phương trình

a, $\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}\frac{x+4}{2001}$

b, $\frac{201-x}{99}+\frac{205-x}{97}+\frac{205-x}{95}+3=0$

c, $\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}$

1

TT

Trần Thanh Phương

6 tháng 7 2019

a) $\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}$

$\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}$

$\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0$

$\Leftrightarrow x+2005=0$

$\Leftrightarrow x=-2005$

b) Sửa đề :

$\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0$

$\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0$

$\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0$

$\Leftrightarrow x=300$

c) $\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}$

$\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1$

$\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}$

$\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0$

$\Leftrightarrow x=2004$

Vậy....

Bài $1:$TÌM $x:$$a,\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}$$b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}$$c,\frac{x+5}{205}+\frac{x+4}{204}+\frac{x+3}{203}=\frac{x+166}{366}+\frac{x+167}{367}+\frac{x+168}{368}$ $d,$ ...

NV

nguyen van huy

19 tháng 7 2016

#Toán lớp 6

1

TN

Thắng Nguyễn

19 tháng 7 2016

a)$\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}$

$\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}$

$2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}$

$\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}$

đề sai

b)$\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1$

$\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}$

$\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0$

$\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0$

$x+2004=0$.Do $\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0$

$x=-2004$

c)$\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1$

$\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}$

$\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0$

$\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0$

$x-200=0$.Do$\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0$

$x=200$

d)chịu

1.Tìm các số thực a, b, c biết:$\frac{5x^2-2}{x^3-3x-2}=\frac{a}{x-2}+\frac{b}{x+1}+\frac{c}{\left(x+1\right)^2}$2.giải các phương trình...

NH

nguyen haili

21 tháng 1 2017

1

NH

nguyen haili

24 tháng 1 2017

Giúp với

Bài 5: Giải các phương trình sau : a) $\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}-6}{5}$ b) $\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}$ c) $\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}$ d) $\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}$ e) $\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}$ ...

HT

Huyền Trang

3 tháng 5 2020

Giải các phương trình sau a) $\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}$ b) $\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)$ c) $\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}$ d) $\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0$ ...

0

NH

nguyễn hoài thu

6 tháng 2 2020

2

AH

Akai Haruma

Giáo viên

7 tháng 2 2020

a)

$\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}$

$\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0$

Dễ thấy: $\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0$

$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$

Do đó $x-23=0\Rightarrow x=23$

b)

PT $\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}$

$\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0$

Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$

$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$

$\Rightarrow x+100=0\Rightarrow x=-100$

AH

Akai Haruma

Giáo viên

7 tháng 2 2020

c)

PT $\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1$

$\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}$

$\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0$

Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$

Do đó $x+2005=0\Rightarrow x=-2005$

d)

PT $\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0$

$\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0$

$\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0$

Dễ thấy $\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0$ hay khác $0$

Do đó $300-x=0\Rightarrow x=300$