1. x(x-3)-(x+2)(x-1)=3 <=> x² - 3x - x² - x + 2 = 3 => 4x = -1 => x = 1/4 

2. 

a) x = 0, x=1 (2 nghiệm, loại)

b) x² + 1 > 0 => x = - 2 (1 nghiệm, chọn b)

c) <=> x(x-3) = 0 => x = 0, x=3 (2 nghiệm, loại)

d) (x-1)² + 2 > 0 => Vô nghiệm (loại)

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

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

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

\(\Leftrightarrow2x=1\)

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

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

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

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

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

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

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

c: =>(x-3)(x²+3x+5)=0

=>x-3=0

hay x=3

d: =>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x-3)(x-4)=0

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

a: =>(5x+3)(x-1)=0

=>x=1 hoặc x=-3/5

b: =>(x-3)(4x-1-5x-2)=0

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

=>x=-3 hoặc x=3

c: =>(x+6)(3x-1+x-6)=0

=>(x+6)(4x-7)=0

=>x=7/4 hoặc x=-6

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

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

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)

1) `x^2+4-2(x-1)=(x-2)^2`

`<=>x^2+4-2x+2=x^2-4x+4`

`<=>-2x+2=-4x`

`<=>2x=-2`

`<=>x=-1`

.

2) ĐKXĐ: `x \ne \pm 3`

`(x+3)/(x-3)-(x-1)/(x+3)=(x^2+4x+6)/(x^2-9)`

`<=>(x+3)^2-(x-1)(x-3)=x^2+4x+6`

`<=>x^2+6x+9-x^2+4x-3=x^2+4x+6`

`<=>10x+6=x^2+4x+6`

`<=>x^2-6x=0`

`<=>x(x-6)=0`

`<=>x=0;x=6`

.

3) ĐKXĐ: `x \ne \pm 3`

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

`<=>(3x-3)-(x+3)=(x+1)(x-3)`

`<=> 2x-6=x^2-2x-3`

`<=>x^2-4x+3=0`

`<=>x^2-x-3x+3=0`

`<=>x(x-1)-3(x-1)=0`

`<=>(x-3)(x-1)=0`

`<=> x=3;x=1`

Vậy...

a: =>x-3=2 hoặc x-3=-2

=>x=5 hoặc x=1

b: =>x²=0

hay x=0

c: =>(3x-5-x+1)(3x-5+x-1)=0

=>(2x-4)(4x-6)=0

=>x=2 hoặc x=3/2

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

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

hay \(x\in\left\{1;-1;4\right\}\)

\(a,\left(x-3\right)^2=4\\\Leftrightarrow\left(x-3\right)^2-2^2=0\\ \Leftrightarrow \left(x-3-2\right).\left(x-3+2\right)=0\\ \Leftrightarrow\left(x-5\right).\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\\\Rightarrow S=\left\{1;5\right\}\\ b,x^2.\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow S=\left\{0\right\}\\ c,\left(3x-5\right)^2-\left(x-1\right)^2=0\\ \Leftrightarrow\left(3x-5-x+1\right).\left(3x-5+x-1\right)=0\\ \Leftrightarrow\left(2x-4\right).\left(4x-6\right)=0\\ \Leftrightarrow2.\left(x-2\right).2.\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow S=\left\{\dfrac{3}{2};2\right\}\)

\(d,\left(x^2-1\right).\left(2x-1\right)=\left(x^2-1\right).\left(x+3\right)\\ \Leftrightarrow\left(x^2-1\right).\left(2x-1-x-3\right)=0\\ \Leftrightarrow\left(x^2-1\right).\left(x-4\right)=0\\ \Leftrightarrow\left(x-1\right).\left(x+1\right).\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=4\end{matrix}\right.\\ \Rightarrow S=\left\{-1;1;4\right\}\)

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

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

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

a)

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

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

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

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

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

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

\(\Leftrightarrow\left(x+1\right)\left(6x^2+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\6x^2+x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(2x-1\right)\left(3x+2\right)=0\end{matrix}\right.\)

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

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

b)

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

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

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

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=\left(\pm1\right)^2\)

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

Vậy \(x\in\left\{1;-1\right\}\)

a) (3x + 2)² - (3x - 2)² = 5x + 38

<=> 6x.4 = 5x + 38 <=> 19x = 38 <=> x = 2

b) 3(x - 2)² + 9(x - 1) = 3(x² + x - 3)

<=> 3x² - 12x + 12 + 9x - 9 = 3x² + 3x - 9

<=> -6x = -12 <=> x = 2

c) (x + 3)² - (x - 3)² = 6x + 8

<=> 2x.6 = 6x + 8 <=> 6x = 8 <=> x = 4/3

d) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

<=> x³ - 3x² + 3x - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> 3x = -21 <=> x = -7

e) (x + 1)(x² - x + 1) - 2x = x(x - 1)(x + 1)

<=> x³ - 1 - 2x = x³ - x

<=> x = -1