c) \(\dfrac{x}{x-2}+\dfrac{x}{x+2}=\dfrac{4x}{x^2-4}.ĐKXĐ:x\ne2;-2\)

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

<=>x²+2x+x²-2x=4x

<=>2x²-4x=0

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

<=>\(\left[{}\begin{matrix}2x=0< =>x=0\\x-2=0< =>x=2\left(loại\right)\end{matrix}\right.\)

Vậy pt trên có nghiệm là S={0}

d) 11x-9=5x+3

<=>11x-5x=9+3

<=>6x=12

<=>x=2

Vậy pt trên có nghiệm là S={2}

e) (2x+3)(3x-4) =0

<=> \(\left[{}\begin{matrix}2x+3=0< =>x=\dfrac{-3}{2}\\3x-4=0< =>x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy pt trên có tập nghiệm là S={\(\dfrac{-3}{2};\dfrac{4}{3}\)}

a) 5x+9 =2x

<=> 5x-2x=9

<=> 3x=9

<=> x=3

Vậy pt trên có nghiệm là S={3}

b) (x+1)(4x-3)=(2x+5)(x+1)

<=> (x+1)(4x-3)-(2x+5)(x+1)=0

<=>(x+1)(2x-8)=0

<=>\(\left[{}\begin{matrix}x+1=0< =>x=-1\\2x-8=0< =>2x=8< =>x=4\end{matrix}\right.\)

Vậy pt trên có tập nghiệm là S={-1;4}

c) 

<=>

<=>x²+2x+x²-2x=4x

<=>2x²-4x=0

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

<=>

Vậy pt trên có nghiệm là S={0}

d) 11x-9=5x+3

<=>11x-5x=9+3

<=>6x=12

<=>x=2

Vậy pt trên có nghiệm là S={2}

e) (2x+3)(3x-4) =0

<=> 

Vậy pt trên có tập nghiệm là S={}

\(a,\left(2x-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(2x-3\right)\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=-2\end{matrix}\right.\\ b,2x-\left(3-5x\right)=4\left(x+3\right)\\ \Leftrightarrow2x-3+5x=4x+12\\ \Leftrightarrow7x-3-4x-12=0\\ \Leftrightarrow3x-15=0\\ \Leftrightarrow x=5\)

\(c,ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}-\dfrac{2}{x+1}=\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}-\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}-\dfrac{11-3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x+1-x+2-11+3x}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow3x-8=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)

a: =>(x-2)(2x+5)=0

=>x-2=0 hoặc 2x+5=0

=>x=2 hoặc x=-5/2

c: \(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)

=>\(\dfrac{2x^2+2x-x^2+x}{x^2-1}=1\)

=>x^2+3x=x^2-1

=>3x=-1

=>x=-1/3

\(a,\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{2;\dfrac{5}{2}\right\}\)

\(c,\Leftrightarrow2x.\left(x+1\right)-x.\left(x-1\right)=\left(x-1\right)\left(x+1\right)\)              ( ĐKXĐ: \(x\ne-1;x\ne1\) )

\(\Leftrightarrow2x^2+2x-x^2+x=x^2-1\\ \Leftrightarrow x^2-x^2+3x=-1\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\)  ( nhận )

Vậy phương trình có tập nghiệm S = \(\left\{-\dfrac{1}{3}\right\}\)

a: =>x-2=0 hoặc x+3=0

=>x=2 hoặc x=-3

b:=>x-7=0 hoặc x+2=0

=>x=7 hoặc x=-2

c: =>4x+2=0 hoặc 3x-4=0

=>x=4/3 hoặc x=-1/2

d: =>2x+1=0 hoặc x-3=0

=>x=3 hoặc x=-1/2

a)

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

`<=> x-2=0` hoặc `x+3=0`

`<=>x=2` hoặc `x=-3`

b)

`(x-7)(2+x)=0`

`<=>x-7=0` hoặc `2+x=0`

`<=>x=7` hoặc `x=-2`

c)

`(4x+2)(3x-4)=0`

`<=>4x+2=0` hoặc `3x-4=0`

`<=>x=-1/2` hoặc `x=4/3`

d)

`(2x+1)(x-3)=0`

`<=>2x+1=0` hoặc `x-3=0`

`<=>x=-1/2` hoặc `x=3`

e)

`(0,1x-3)(x+0,5)=0`

`<=>0,1x-3=0` hoặc `x+0,5=0`

`<=>x=30` hoặc `x=-0,5`

f)

`(0,2x-0,4)(0,1x+0,7)=0`

`<=>0,2x-0,4=0` hoặc `0,1x+0,7=0`

`<=>x=2` hoặc `x=-7`

Câu 1 : 

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

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

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

Bài 3: 

b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)

hay \(x\in\left\{0;-1\right\}\)

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

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

Bài 1:

a) 5(x-3)-4=2(x-1)

\(\Leftrightarrow5x-15-4=2x-2\)

\(\Leftrightarrow5x-19-2x+2=0\)

\(\Leftrightarrow3x-17=0\)

\(\Leftrightarrow3x=17\)

\(\Leftrightarrow x=\frac{17}{3}\)

Vậy: \(x=\frac{17}{3}\)

b) 5-(6-x)=4(3-2x)

\(\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow-1+x-12+8x=0\)

\(\Leftrightarrow-13+9x=0\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\frac{13}{9}\)

Vậy: \(x=\frac{13}{9}\)

c) (3x+5)(2x+1)=(6x-2)(x-3)

\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)

\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)

\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)

\(\Leftrightarrow33x-1=0\)

\(\Leftrightarrow33x=1\)

\(\Leftrightarrow x=\frac{1}{33}\)

Vậy: \(x=\frac{1}{33}\)

d) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)

\(\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8\)

\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)

\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)

\(\Leftrightarrow12x-12=0\)

\(\Leftrightarrow x=1\)

Vậy:x=1

Bài 2:

a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)

\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)

\(\Leftrightarrow4x-10x-15x-3x+60=0\)

\(\Leftrightarrow-24x+60=0\)

\(\Leftrightarrow-24x=-60\)

\(\Leftrightarrow x=\frac{5}{2}\)

Vậy: \(x=\frac{5}{2}\)

b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)

\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)

\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)

\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)

\(\Leftrightarrow-3x=0\)

\(\Leftrightarrow x=0\)

Vậy: x=0

c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)

\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)

\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)

\(\Leftrightarrow15x-15-2x-2-10x+65=0\)

\(\Leftrightarrow3x+48=0\)

\(\Leftrightarrow3x=-48\)

\(\Leftrightarrow x=-16\)

Vậy: x=-16

d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)

\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)

\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)

\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)

\(\Leftrightarrow-13x+143=0\)

\(\Leftrightarrow-13x=-143\)

\(\Leftrightarrow x=11\)

Vậy: x=11

e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)

\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)

\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)

\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)

\(\Leftrightarrow45x-18-24-28x+60x-420=0\)

\(\Leftrightarrow77x-462=0\)

\(\Leftrightarrow77x=462\)

\(\Leftrightarrow x=6\)

Vậy:x=6

Bài 3:

a) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)

Vì \(2\ne0\)

nên \(\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)

b) \(\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)

c) \(\left(2x+1\right)\left(x^2+2\right)=0\)

Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(2x+1=0\)

\(\Leftrightarrow2x=-1\)

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

Vậy: \(x=\frac{-1}{2}\)

d) \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

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

Ta có: \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)

Ta lại có \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)

Ta có: \(4\ne0\)(4)

Từ (3) và (4) suy ra

2x-1=0

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy: \(x=\frac{1}{2}\)

Bài 4:

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

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

\(\Leftrightarrow2x^2-x-6=x^2-3x+2\)

\(\Leftrightarrow2x^2-x-6-x^2+3x-2=0\)

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

\(\Leftrightarrow x^2+2x+1-9=0\)

\(\Leftrightarrow\left(x+1\right)^2-3^2=0\)

\(\Leftrightarrow\left(x+1-3\right)\left(x+1+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-4\right\}\)

b) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)

\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)

Vì \(3\ne0\)

nên \(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{0;4\right\}\)

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

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)

d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)

\(\Leftrightarrow-8x^2+40x-32=0\)

\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)

\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)

Vì \(-8\ne0\)

nên \(x^2-5x+4=0\)

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

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{1;4\right\}\)

e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)

\(\Leftrightarrow7x^2+58x+115=0\)

\(\Leftrightarrow7x^2+23x+35x+115=0\)

\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)

\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)

Bài 5:

a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\x=-1\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)

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

\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow2x^2-2x+x^2+2x-3=0\)

\(\Leftrightarrow3x^2-3=0\)

\(\Leftrightarrow3\left(x^2-1\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)

Vì \(3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-1\right\}\)

c) \(x^4+x^3+x+1=0\)

\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)

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

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)

Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)

Từ (5) và (6) suy ra

\(\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy: x=-1

ko khó đâu, chủ yếu nhát làm

Câu 1:

a.5.(x-3)-4=2.(x-1)

⇔5x-15-4=2x-2

⇔ 5x-2x=-2+19

⇔ 3x=17

⇔ x=17/3

b. 5-(6-x)=4.(3-2x)

⇔ x-1=12-8x

⇔ x+8x=12+1

⇔ x=13/9

c.(3x+5).(2x+1)=(6x-2).(x-3)

⇔ 6x² + 3x+10x+5=6x²-18x-2x+6

⇔ (6x²-6x²)+(13x+20x)=6-5

⇔ 33x=1

⇔x=1/33

d.(x+2)²+2.(x-4)=(x-4).(x-2)

⇔x²+4x+4+2x-8=x²-2x-4x+8

⇔(x²-x²)+(6x+6x)=8+8-4

⇔12x=12

⇔ x=1