( 2x - 3 )( x + 1 ) - 2x² + 6x = 0

<=> 2x² - x - 3 - 2x² + 6x = 0

<=> 5x - 3 = 0

<=> 5x = 3

<=> x = 3/5

( x² - x + 1 )( x - 3 ) - x³ + 4x² = 0

<=> x³ - 4x² + 4x - 3 - x³ + 4x² = 0

<=> 4x - 3 = 0

<=> 4x = 3

<=> x = 3/4

( x² - 2 )( x² + 2 ) - x⁴ - 2x + 5 = 0

<=> ( x² )² - 4 - x⁴ - 2x + 5 = 0

<=> x⁴ + 1 - x⁴ - 2x = 0

<=> 1 - 2x = 0

<=> 2x = 1

<=> x = 1/2

( x - 3 )( x² - 3x + 2 ) - ( x² - 2x - 7 )( x - 2 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - ( x³ - 4x² - 3x + 14 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - x³ + 4x² + 3x - 14 + 2x² - 2x = 0

<=> 12x - 20 = 0

<=> 12x = 20

<=> x = 20/12 = 5/3

a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)

\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)

b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)

\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)

c ; d tương tự nhé ! 

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

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

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

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

=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0

=>x=1/2 hoặc x=-3 hoặc x=2

2)x^3 + x^2 + x + 1 = 0

=>.x^2(x+1)+(x+1)=0

=>(x^2+1)(x+1)=0

=>x^2+1=0 hoặc x+1=0 

=>                      x =-1

3) 2x(x-3)+5(x-3) =0    

=>(2x+5)(x-3)=0

=>2x+5=0 hoặc x-3=0

=>x=-5/2 hoặc x=3

4)x(2x-7)-(4x-14)=0

=> (x-2)(2x-7)=0

=> x-2 =0 hoặc 2x-7=0

=>x=2 hoặc x=7/2

5)2x^3+3x^2+2x+3=0

=>x^2(2x+3)+2x+3=0

=>(x^2+1)(2x+3)=0

=>x^2+1=0 hoặc 2x+3=0

=>                      x =-3/2

x = 3/2 đó mình chắc chắn 100 %

X= 3/2 nha bạn, chắc chắn đó.

a: \(3\left(x-3\right)-6x=0\)

=>\(3x-9-6x=0\)

=>-3x-9=0

=>3x+9=0

=>3x=-9

=>\(x=-\dfrac{9}{3}=-3\)

b: Đề thiếu vế phải rồi bạn

c: \(2\left(x-3\right)+3x=9\)

=>2x-6+3x=9

=>5x-6=9

=>5x=6+9=15

=>x=15/5=3

d: \(x\left(x-11\right)+2\left(x-11\right)=0\)

=>\(\left(x-11\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)

e: \(x\left(x+2\right)+8=x^2\)

=>\(x^2+2x+8=x^2\)

=>2x+8=0

=>2x=-8

=>x=-8/2=-4

f: \(8\left(x+1\right)+2x=-2\)

=>\(8x+8+2x=-2\)

=>10x=-2-8=-10

=>\(x=-\dfrac{10}{10}=-1\)

g: 12-3(x+2)=0

=>3(x+2)=12

=>x+2=12/3=4

=>x=4-2=2

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

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

\(\Leftrightarrow-x^3+5x^2-5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2-\frac{5}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x-\frac{5}{2}\right)^2=\frac{5}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x-\frac{5}{2}=\frac{\sqrt{5}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{5+\sqrt{5}}{2}\\x=\frac{5-\sqrt{5}}{2}\end{cases}}\)

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

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

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

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

=> \(x\in\left\{0;\frac{5+\sqrt{5}}{2};\frac{5-\sqrt{5}}{2}\right\}\)

b) \(\left(2x-5\right)\left(x+3\right)-\left(x-1\right)\left(2x+3\right)=0\)

\(\Leftrightarrow2x^2+x-15-2x^2-x+3=0\)

\(\Leftrightarrow-12=0\left(vn\right)\)

c) \(\left(x-2\right)\left(x^2+2x+8\right)-x^3-2x+1=0\)

\(\Leftrightarrow x^3+4x-16-x^3-2x+1=0\)

\(\Leftrightarrow2x=15\)

\(\Rightarrow x=\frac{15}{2}\)

a) x( x - 2 ) - x( x - 1 )( x - 3 ) = 0

<=> x² - 2x - x( x² - 4x + 3 ) = 0

<=> x² - 2x - x³ + 4x² - 3x = 0

<=> -x³ + 5x² - 5x = 0

<=> -x( x² - 5x + 5 ) = 0

<=> \(\orbr{\begin{cases}-x=0\\x^2-5x+5=0\end{cases}}\)

+) -x = 0 => x = 0

+) x² - 5x + 5 = 0 (*)

\(\Delta=b^2-4ac=\left(-5\right)^2-4\cdot1\cdot5==25-20=5\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{5+\sqrt{5}}{2}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{5-\sqrt{5}}{2}\end{cases}}\)

Vậy S = { \(0;\frac{5\pm\sqrt{5}}{2}\)}

b) ( 2x - 5 )( x + 3 ) - ( x - 1 )( 2x + 3 ) = 0

<=> 2x² + x - 15 - ( 2x² + x - 3 ) = 0

<=> 2x² + x - 15 - 2x² - x + 3 = 0

<=> -12 = 0 ( vô lí )

Vậy phương trình vô nghiệm

c) ( x - 2 )( x² + 2x + 8 ) - x³ - 2x + 1 = 0

<=> x³ + 4x - 16 - x³ - 2x + 1 = 0

<=> 2x - 15 = 0

<=> 2x = 15

<=> x = 15/2

2: \(3x\left(x-4\right)+2x-8=0\)

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

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

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

3: 4x(x-3)+x²-9=0

=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(4x+x+3\right)=0\)

=>\(\left(x-3\right)\left(5x+3\right)=0\)

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

4: \(x\left(x-1\right)-x^2+3x=0\)

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

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

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

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

vậy ...

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

    \(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: Ta có: \(x^4-2x^3+5x^2-10x=0\)

\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)

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

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

b:Ta có: \(\left(3x+5\right)^2=\left(2x-2\right)^2\)

\(\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\)

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

\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2