2. Tìm x biết: a)$\dfrac{2}{\left(x 2\right)\left(x 4\right)}$ $\dfrac{4}{\left(x 4\right)\left(x 8\right)}$ $\dfrac{6}{\left(x 8\right)\left(x 14\right)}$ = \(\dfrac{x}{\left(x 2\right)\le...

2. Tìm x biết:a)2(x+2)(x+4)\dfrac{2}{\left(x+2\right)\left(x+4\right)} + 4(x+4)(x+8)\dfrac{4}{\left(x+4\right)\left(x+8\right)} + 6(x+8)(x+14)\dfrac{6}{\left(x+8\right)\left(x+14\right)} = x(x+2)(x+14)\dfrac{x}{\left(x+2\right)\left(x+14\right)}b)x2023\dfrac{x}{2023} + x+12022\dfrac{x+1}{2022}+ x+22021\dfrac{x+2}{2021} +...+ x+20221\dfrac{x+2022}{1} + 2023 = 0.Gíup mình giải 2 bài này với!Cảm ơn các bạn rất...

2

NN

Nguyễn Ngọc Anh Minh

VIP

14 tháng 8 2023

a/

$VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=$

$=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=$

$=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}$

$\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)$

$\Rightarrow x=12$

Đúng(1)

WT

when the imposter is sus

14 tháng 8 2023

$\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0$

$\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0$

$x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0$

$x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023$

$x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}$

$x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}$

$x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023$

Vậy x = 2023

Đúng(1)

JW

Jackson Williams

14 tháng 8 2023

0

NL

Nguyễn Lê Phương Linh

16 tháng 9 2023

$\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{4}{\left(x+4\right)\left(x+8\right)}+\dfrac{6}{\left(x+8\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}$

1

AH

Akai Haruma

Giáo viên

16 tháng 9 2023

Lời giải:

Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$

Đề bài

$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow 12=x$ (thỏa mãn)

Đúng(1)

KY

Khinh Yên

13 tháng 7 2018

Tìm x biết

a)$\dfrac{2}{\left(x+2\right)\times\left(x+4\right)}+\dfrac{4}{\left(x+4\right)\times\left(x+8\right)}+\dfrac{6}{\left(x+8\right)\times\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\times\left(x+14\right)}$

1

AH

Akai Haruma

Giáo viên

13 tháng 7 2018

Lời giải:

PT $\Leftrightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Leftrightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow x=12$ (thỏa mãn)

Vậy......

2S

2012 SANG

30 tháng 8 2023

$\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3$

$\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}$

$\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4$

$\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}$

$\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6$

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

30 tháng 8 2023

6:ĐKXĐ: x>=0; x<>1/25

BPT=>$\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0$

=>$\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0$

=>$\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0$

=>$\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}$

=>$\dfrac{1}{25}< x< =\dfrac{25}{324}$

7:

ĐKXĐ: x>=0

BPT $\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1$

=>$\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0$

=>$\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0$

=>$-\sqrt{x}-2>=0$(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT $\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0$

=>$\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0$

=>$\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0$

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9:

ĐKXĐ: x>=0

$BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}$

=>$\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0$

=>$\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0$

=>$\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0$(vô lý)

10:

ĐKXĐ: x>=0; x<>1/49

$BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0$

=>$\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0$

=>$\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0$

=>$\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0$

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

2S

2012 SANG

30 tháng 8 2023

câu 9 nhầm đề bài r bạn

QH

Quang Huy Điền

5 tháng 12 2017

a) $8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2$

b) $\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=6$

c) $\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=840$

#Toán lớp 8

0

NT

Nguyễn Thu Huyền

1 tháng 9 2023

Tìm...

Tìm x a, $\dfrac{\left(x+2\right)^2}{2}$ + $\dfrac{\left(1+2x\right)^2}{4}$ + $\dfrac{\left(1-2x\right)^2}{8}$ – (1 + x)2 = 0b, $\dfrac{\left(x+1\right)^2}{2}$ - $\dfrac{\left(1-2x\right)^2}{3}$ + $\dfrac{\left(1+2x\right)^2}{4}$ - $\dfrac{\left(5-x\right)^2}{6}$= 0c, (3 + x)3 – 3x2(x + 4) + (x + 2)3 = (1 – x)3 –...

3

T

Toru

1 tháng 9 2023

$a,3-x=x+1,8$

$\Rightarrow-x-x=1,8-3$

$\Rightarrow-2x=-1,2$

$\Rightarrow x=0,6$

$b,2x-5=7x+35$

$\Rightarrow2x-7x=35+5$

$\Rightarrow-5x=40$

$\Rightarrow x=-8$

$c,2\left(x+10\right)=3\left(x-6\right)$

$\Rightarrow2x+20=3x-18$

$\Rightarrow2x-3x=-18-20$

$\Rightarrow-x=-38$

$\Rightarrow x=38$

$d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x$

$\Rightarrow8x-3+1=1+6x+x$

$\Rightarrow8x-3=7x$

$\Rightarrow8x-7x=3$

$\Rightarrow x=3$

$e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x$

$\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}$

$\Rightarrow-2x=\dfrac{10}{9}$

$\Rightarrow x=-\dfrac{5}{9}$

Đúng(3)

T

Toru

1 tháng 9 2023

$g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}$

$\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}$

$\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}$

$\Rightarrow x=\dfrac{16}{3}$

$h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)$

$\Rightarrow x-4=5-x$

$\Rightarrow x+x=5+4$

$\Rightarrow2x=9$

$\Rightarrow x=\dfrac{9}{2}$

$k,7x^2-11=6x^2-2$

$\Rightarrow7x^2-6x^2=-2+11$

$\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.$

$m,5\left(x+3\cdot2^3\right)=10^2$

$\Rightarrow5\left(x+24\right)=100$

$\Rightarrow x+24=20$

$\Rightarrow x=-4$

$n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}$

$\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}$

$\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}$

$\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0$

$\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}$

#$Urushi\text{☕}$

Đúng(3)

HA

Hikaru Akira

12 tháng 9 2021

#Toán lớp 8

3

NL

Nguyễn Lê Phước Thịnh

12 tháng 9 2021

a: ta có: $\dfrac{\left(x+2\right)^2}{2}+\dfrac{\left(2x+1\right)^2}{4}+\dfrac{\left(2x-1\right)^2}{8}-\left(x+1\right)^2=0$

$\Leftrightarrow4\left(x^2+4x+4\right)+2\left(4x^2+4x+1\right)+4x^2-4x+1-8\left(x+1\right)^2=0$

$\Leftrightarrow4x^2+16x+16+8x^2+8x+2+4x^2-4x+1-8\left(x^2+2x+1\right)=0$

$\Leftrightarrow16x^2+20x+19-8x^2-16x-8=0$

$\Leftrightarrow8x^2+4x+11=0$

$\text{Δ}=4^2-4\cdot8\cdot11=-336< 0$

Vì Δ<0 nên phương trình vô nghiệm

Đúng(3)

AH

Akai Haruma

Giáo viên

13 tháng 9 2021

b.

PT $\Leftrightarrow \frac{x^2+2x+1}{2}-\frac{4x^2-4x+1}{3}+\frac{4x^2+4x+1}{4}-\frac{x^2-10x+25}{6}=0$

$\Leftrightarrow \left(\frac{x^2+2x+1}{2}+\frac{4x^2+4x+1}{4}\right)-\left(\frac{4x^2-4x+1}{3}+\frac{x^2-10x+25}{6}\right)=0$

$\Leftrightarrow \frac{6x^2+8x+3}{4}-\frac{9x^2-18x+27}{6}=0$

$\Leftrightarrow \frac{3(6x^2+8x+3)-2(9x^2-18x+27)}{12}=0$

$\Leftrightarrow 5x-\frac{15}{4}=0$

$\Leftrightarrow x=\frac{3}{4}$

N

nguyenviethung

3 tháng 10 2023

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}$+$\dfrac{2}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}$

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 12 2023

Sửa đề:

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}$

ĐKXĐ: $x\notin\left\{1;3;8;20\right\}$

PT=>$-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}$

=>$-\dfrac{1}{x-4}=-\dfrac{3}{4}$

=>$x-1=\dfrac{4}{3}$

=>$x=\dfrac{4}{3}+1=\dfrac{7}{3}$(nhận)

Giải PT: a, $\dfrac{x^2+x+1}{x^2+x+2}+\dfrac{x^2+x+2}{x^2+x+3}=\dfrac{7}{6}$ b, $\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}$ c, $8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2$ Help me!!! Mk cần...

HA

Huyền Anh Kute

11 tháng 3 2018

#Toán lớp 8

4

HH

hattori heiji

11 tháng 3 2018

đkxđ với mọi x

đặt a=x²+x+1

$\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}$

<=> $\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}$

=> 6a(a+2) +6(a+1)² =7(a+1)(a+2)

<=> 6a²+12a +6a² +12a+6 =a²+21a+14

<=> 12a² -a²+24a-21a+6-14=0

<=> 11a²+3a-8=0

<=> 11a² +11a-8a-8=0

<=> (11a² +11a)-(8a+8)=0

<=> 11a(a+1)-8(a+1)=0

<=> (a+1)(11a-8)=0

=> a=-1 và a=$\dfrac{8}{11}$

thay a=x²+x+1 ta đc

x²+x+1=-1

<=> x²+x+2 =0 (vô nghiệm)

và x²+x+$\dfrac{3}{11}$ =0(vô nghiệm )

vậy pt trên vô nghiệm

PK

Phùng Khánh Linh

12 tháng 3 2018

c) $8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)$ĐKXĐ : x # 0

( 2) <=> $8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2$

$< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2$

$< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2$

$< =>16=\left(x+4\right)^2$

<=> x² + 8x = 0

<=> x( x + 8) = 0

<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )

Vậy,....