\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)

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

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)

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

\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)

\(\Leftrightarrow0x=0\)

Phương trình có vô số nghiệm , trừ x = -1,x = 3

Vậy ...

\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)

\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)

\(\Leftrightarrow12x+1< 36x+4-24x-3\)

\(\Leftrightarrow12x+1< 12x+1\)

\(\Leftrightarrow12x-12x< 1-1\)

\(\Leftrightarrow0x< 0\)

Vậy S = {x | x \(\in R\)}

\(\Leftrightarrow2x^2+3x-4x-6-2\left(x^2-1\right)=6\)

\(\Leftrightarrow2x^2-x-6-2x^2+2-6=0\)

=>x+10=0

hay x=-10

\(\left(x-2\right)\left(2x+3\right)-2\left(x-1\right)\left(x+1\right)=6\\ \Rightarrow2x^2-4x+3x-6-2x^2+2-6=0\\ \Rightarrow-x-10=0\\ \Rightarrow-x=10\\ \Rightarrow x=-10\)

\(\left(x-2\right)\left(2x+3\right)-2\left(x-1\right)\left(x+1\right)=6\)

\(\Leftrightarrow2x^2-x-6-2\left(x^2-1\right)=6\Leftrightarrow-x=10\Leftrightarrow x=-10\)

1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)

\(ĐK:x\ne3;-2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)+x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow\left(x+2\right)\left(x+2\right)+x\left(x-3\right)=x^2+6\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)

\(\Leftrightarrow\)    \(\left(x+1\right)^2+x-1=x^2+4\) hoặc   \(\left(x+1\right)^2+1-x=x^2+4\)

Xét \(\left(x+1\right)^2+x-1=x^2+4\)

\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)

\(\Leftrightarrow3x-4=0\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

Xét \(\left(x+1\right)^2+1-x=x^2+4\)

\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)

2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)

\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)

\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)

\(\Leftrightarrow-x< 4\)

\(\Leftrightarrow x>4\)

Vậy \(S=\left\{x|x>4\right\}\)

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.

a: =>3x=3

=>x=1

b: =>12x-2(5x-1)=3(8-3x)

=>12x-10x+2=24-9x

=>2x+2=24-9x

=>11x=22

=>x=2

c: =>2x-3(2x+1)=x-6x

=>-5x=2x-6x-3=-4x-3

=>-x=-3

=>x=3

d: =>2x-5=0 hoặc x+3=0

=>x=5/2 hoặc x=-3

e: =>x+2=0

=>x=-2

Mình khuyên bạn thế này : 

Bạn nên tách những câu hỏi ra 

Như vậy các bạn sẽ dễ giúp

Và cũng có nhiều bạn giúp hơn !

Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

b) ( x - 2 )² + ( x - 2 )( x - 3 ) = 0

<=> ( x - 2 )( x - 2 + x - 3 ) = 0

<=> ( x - 2 )( 2x - 5 ) = 0

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

<=> x( x - 2 ) - 3( x - 2 ) = 0

<=> ( x - 2 )( x - 3 ) = 0

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

Bài 2.

a) \(\frac{x}{x+1}-1=\frac{3}{2}x\)

ĐKXĐ : x khác -1

<=> \(\frac{x}{x+1}-\frac{x+1}{x+1}=\frac{3}{2}x\)

<=> \(\frac{-1}{x+1}=\frac{3x}{2}\)

=> 3x( x + 1 ) = -2

<=> 3x² + 3x + 2 = 0

Vi 3x² + 3x + 2 = 3( x² + x + 1/4 ) + 5/4 = 3( x + 1/2 )² + 5/4 ≥ 5/4 > 0 ∀ x

=> phương trình vô nghiệm

b) \(\frac{4x}{x-2}-\frac{7}{x}=4\)

ĐKXĐ : x khác 0 ; x khác 2

<=> \(\frac{4x^2}{x\left(x-2\right)}-\frac{7x-14}{x\left(x-2\right)}=\frac{4x^2-8x}{x\left(x-2\right)}\)

=> 4x² - 7x + 14 = 4x² - 8x

<=> 4x² - 7x - 4x² + 8x = -14

<=> x = -14 ( tm )

Vậy phương trình có nghiệm x = -14

1)  \(3x-2x+6=6\Leftrightarrow x=0\)

2) \(4\left(2x-1\right)-12x-12=3\left(x+2\right)\)

\(\Leftrightarrow8x-4-12x-12-3x-6=0\)

\(\Leftrightarrow7x=-22\Leftrightarrow x=\dfrac{-22}{7}\)

3, \(\left(x-1\right)2=9\left(x+1\right)2\)

\(\Leftrightarrow2x-2\)    \(=18x+18\)

\(\Leftrightarrow2x-18x=18+2\)

\(\Leftrightarrow-16x\)       \(=20\)

\(\Leftrightarrow x\)             \(=\dfrac{-5}{4}\)      

                   Vậy pt đã cho có tập nghiệm là S= \(\left\{\dfrac{-5}{4}\right\}\)

4, \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\) ( ĐKXĐ : \(x\ne\pm1\) )

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

\(\Rightarrow x^2-3x-4+x^2+3x-4=2x^2-2\)

\(\Leftrightarrow2x^2-8-2x^2+2=0\)

\(\Leftrightarrow0\)                           \(=6\)  ( Vô lí )

                        Vậy pt đã cho vô nghiệm

\(x-5=\frac{1}{3\left(x+2\right)}\left(đkxđ:x\ne-2\right)\)

\(< =>3\left(x-5\right)\left(x+2\right)=1\)

\(< =>3\left(x^2-3x-10\right)=1\)

\(< =>x^2-3x-10=\frac{1}{3}\)

\(< =>x^2-3x-\frac{31}{3}=0\)

giải pt bậc 2 dễ r

\(\frac{x}{3}+\frac{x}{4}=\frac{x}{5}-\frac{x}{6}\)

\(< =>\frac{4x+3x}{12}=\frac{6x-5x}{30}\)

\(< =>\frac{7x}{12}=\frac{x}{30}< =>12x=210x\)

\(< =>x\left(210-12\right)=0< =>x=0\)

\(3x\left(2x-3\right)-3\left(3+2x^2\right)=0\)

\(< =>6x^2-9x-9-6x^2=0\)

\(< =>-9x-9=0< =>9x+9=0\)

\(< =>x=-\frac{9}{9}=-1\)

\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\left(x\ne-2;x\ne3\right)\\ < =>\dfrac{9x-2}{x^2-3x+2x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{x\left(x-3\right)+2\left(x-3\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\\ < =>\dfrac{9x-2}{\left(x-3\right)\left(x+2\right)}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)

suy ra: \(9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)

\(< =>9x-2+2x^2-6x-\left(x^2+2x-x-2\right)=x^2+2x-3x-6\)

\(< =>9x-2+2x^2-6x-x^2-2x+x+2=x^2-x-6\)

\(< =>2x^2-x^2-x^2+9x-6x-2x+x+x=6+2-2\)

\(< =>3x=6\\ < =>x=2\left(tm\right)\)

ĐKXĐ: \(x\ne\left\{-2;3\right\}\)

\(\dfrac{9x-2}{x^2-x-6}+\dfrac{2x}{x+2}-\dfrac{x-1}{x-3}=1\)

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

\(\Leftrightarrow9x-2+2x\left(x-3\right)-\left(x-1\right)\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)

\(\Leftrightarrow9x-2+2x^2-6x-x^2-x+2=x^2-x-6\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-2\left(loại\right)\)

Vậy: PT vô nghiệm.