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 %

( 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 ! 

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

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

\(\Leftrightarrow2x=15\)

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

1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42

a/

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

\(\Leftrightarrow-6x+6=0\)

\(\Leftrightarrow x=1\)

b/

\(\Leftrightarrow2x^2-4x-2x^2-6x=0\)

\(\Leftrightarrow-10x=0\)

\(\Leftrightarrow x=0\)

c/

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

\(\Leftrightarrow3x\left(2x+3\right)=0\)

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

c/

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(9y^2+30y+25\right)=0\)

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

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

\(\Leftrightarrow x=y=-\frac{5}{3}\)

d/

\(\Leftrightarrow4x^2-4x+1+4x^2+4x+1-2\left(4x^2-2x-2\right)+x=12\)

\(\Leftrightarrow8x^2+x+2-8x^2+4x+4=12\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)

\(a.2x-3=4x+6\) 

\(\Leftrightarrow2x-3-4x-6=0\) 

\(\Leftrightarrow-2x-9=0\)

\(\Leftrightarrow x=\dfrac{9}{2}\)

\(S=\left\{\dfrac{9}{2}\right\}\) 

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

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

\(\Leftrightarrow2x^2+2=0\)

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

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

\(S=\left\{0,-1\right\}\) 

Mấy câu khác bn gửi lại đc ko tại mik chx hiểu lắm

a: =>-2x=9

=>x=-9/2

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

=>x(2x+2)=0

=>x=0 hoặc x=-1

\(a,2x-3=4x+6\)

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

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\dfrac{9}{2}\)

\(b,\) Ghi vậy mình không làm được.

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

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

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

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

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

\(d,\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}=0\left(dkxd:x\ne-1;x\ne3\right)\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x-3\right)-2.2}{2\left(x+1\right)\left(x-3\right)}=0\)

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

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

Vậy \(S=\left\{1\right\}\)

dòng 4 từ dưới lên là ⇒ chứ ko phải ⇔ cj ơi=))

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

=\(\frac{\left(1-x\right)\left(5x+3\right)}{\left(3x-7\right)\left(x-1\right)}=\frac{\left(1-x\right)\left(5x+3\right)}{\left(7-3x\right)\left(1-x\right)}=\frac{\left(5x+3\right)}{\left(7-3x\right)}\)

1, x(x - 5) - 4x + 20 = 0

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

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

=> x - 4 = 0 hoặc x - 5 = 0

=> x = 4 hoặc x = 5

=> x thuộc {4; 5}

2, 3(x + 1) + x(x + 1) 

= (3 + x)(x + 1)

3, 2x³ + x = 0

=> x(2x² + 1) = 0

=> x = 0 hoặc 2x² + 1 = 0

=> x = 0 hoặc 2x² = -1

=> x = 0 hoặc x² = -1/2 (vô lí vì x² > hoặc = 0 với mọi x)

=> x = 0

4, x³ - 16x = 0

=> x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

=> x = 0 hoặc x² = 16

=> x = 0 hoặc x = 4 hoặc x = -4

=> x thuộc {-4; 0; 4}

5, x² + 6x = -9

=> x² + 6x + 9 = 0

=> x² + 2.3.x + 3² = 0

=> (x + 3)² = 0

=> x + 3 = 0

=> x = -3

6, x⁴ - 2x³ + 10x² - 20x = 0

=> x²(x² + 10) - 2x(x² + 10) = 0

=> (x² + 2x)(x² + 10) = 0

=> x(x +2)(x² + 10) = 0

-TH1: x = 0

-TH2: x + 2 = 0 => x = -2

-TH3: x² + 10 = 0 => x² = -10 (vô lí vì x² > hoặc = 0 với mọi x)

=> x thuộc {0; -2}

7, (2x - 3)² = (x + 5)²

-TH1: 2x - 3 = x + 5

=> x = 8

- TH2: - 2x + 3 = x + 5

=> -3x = 2

=> x = \(\frac{-2}{3}\)

- TH3: 2x - 3 = - x - 5

=> 3x = -2

=> x = \(\frac{-2}{3}\)

- TH4: - 2x + 3 = - x - 5

=> -x = -8

=> x = 8`

=> x thuộc {\(\frac{-2}{3}\); 8}