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7 tháng 5 2022

ĐKXĐ: x≠0, x≠2.

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

<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}\) - \(\dfrac{x-2}{x\left(x-2\right)}\)\(\dfrac{2}{x\left(x-2\right)}\)

=> x (x+2) - x + 2 = 2

<=> x^2+2x-x+2=2

<=> x^2+2x-x=2-2

<=> x^2+x=0

<=> x(x+1)=0

<=>x=0 hoặc x+1=0

<=> x=0( ko thỏa mãnĐKXĐ)          x=-1( thỏa mãn ĐKXĐ)

Vậy phương trình có tập nghiệm S={-1} 

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

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,....

\(\Leftrightarrow\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{x\left(x+1\right)}=-1\left(đkxđ:x\ne\pm1;0;2;3\right)\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x}-\dfrac{1}{x+1}=-1\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x+1}=-1\)

\(\Leftrightarrow\dfrac{4}{x^2-2x-3}=-1\)

\(\Leftrightarrow x^2-2x-3=-4\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\left(loai\right)\)

Vậy không có giá trị x thỏa mãn

11 tháng 6 2021

ĐK: `x \ne \pm 3`

`(2(9+2x))/(x^2-9)=2/(x-3)-1/(x+3)`

`<=>2(2x+9)=2(x+3)-(x-3)`

`<=>4x+18=2x+6-x+3`

`<=>4x+18=x+9`

`<=>3x=-9`

`x=-3 (L)`

Vậy `S=∅`.

11 tháng 6 2021

\(\dfrac{2\left(9+2x\right)}{x^2-9}=\dfrac{2}{x-3}-\dfrac{1}{x+3}\)

\(\Leftrightarrow\dfrac{18+4x}{\left(x-3\right)\left(x+3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow18+4x-2x-6+x-3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3\left(x+3\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

\(S=\left\{-3\right\}\)

 

10 tháng 3 2017

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)

ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)

\(\Leftrightarrow3x^2+15x-378=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)

10 tháng 3 2017

@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T

a: =>4x-6-9=5-3x-3

=>4x-15=-3x+2

=>7x=17

hay x=17/7

b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)

=>2/3x+21/3x=4/5+2+1/4=61/20

=>23/3x=61/20

=>3x=23:61/20=460/61

hay x=460/183

13 tháng 12 2021

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)