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

⇔ \(x^3-2x^2+x^2-2x=0\)

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

⇔\(\left(x-2\right)\left(x^2-x\right)=0\)

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

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

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

Vậy phương trình có tập nghiệm S = \(\left\{0,2,-1\right\}\)

a) 2x-7=11x+11

<=> 2x-11x=11+7

<=> -9x=17

<=> x= -17/9

b) 2011x -4 =x+6

<=> 2011x-x=6+4

<=> 2010x=10

<=> x=10/2010

<=> x=1/201

c) 5(2x-3)-2(3x-5)=0

<=> 10x-15-6x+10=0

<=> 10x-6x=15-10

<=>4x=5

<=> x=5/4

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\(\left(x+2\right)^3-16\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left[\left(x+2\right)^2-16\right]=0\)

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

\(\Rightarrow\left(x+2\right)\left(x-2\right)\left(x+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\\x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\\x=-6\end{matrix}\right.\)

Vậy \(S=\left\{-2;2;-6\right\}\)

\(2x^3-6x^2+12x-8=0\)

\(\Rightarrow2x^3-2x^23+3.2^2-2^3=0\)

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

\(\Rightarrow x-2=0\)

\(\Rightarrow x=2\)

`2x^3 +6x^2 =x^2 +3x`

`<=> 2x^3 +6x^2 -x^2 -3x=0`

`<=> 2x^3 +5x^2 -3x=0`

`<=> x(2x^2 +5x-3)=0`

`<=> x(2x^2 +6x-x-3)=0`

`<=> x[2x(x+3)-(x+3)]=0`

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

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

b)

`(2+x)^2 -(2x-5)^2=0`

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

`<=> (-x+7)(3x-3)=0`

\(< =>\left[{}\begin{matrix}-x+7=0\\3x-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

`a) 2x^3 + 6x^2 = x^2 + 3x`

`=> 2x^3 + 6x^2 - x^2 - 3x = 0`

`=> 2x^3 + 5x^2 - 3x = 0`

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

`=> x (2x^2 + 6x - x - 3) = 0`

`=> x [(2x^2 + 6x) - (x+3)] = 0`

`=> x [2x(x+3) - (x+3)] = 0`

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

`=> x = 0` hoặc `2x - 1 = 0` hoặc `x + 3 = 0`

`=> x = 0` hoặc `x = 1/2` hoặc `x = -3`

`b) (2+x)^2 - (2x-5)^2 = 0`

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

`=> (3x - 3)(7-x) = 0`

`=> 3x - 3 = 0` hoặc `7 - x = 0`

`=> x = 1` hoặc `x = 7`

a) Ta có: \(x^3+x^2+x+1=0\)

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

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

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

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

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

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

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

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

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

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

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

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

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

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

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

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

Vậy: S={2;-3}

Giải  phương trình sau

8=11x+6

⇒11x =2

⇒  x = 2/11

Vậy x =2/11

nực cười c/h

a) x² - 4 = 0

x² = 4

x = 2 hoặc x = -2

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

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

X + 5 = 0 hoặc 2x - 3 = 0

*) x + 5 = 0

x = -5

*) 2x - 3 = 0

2x = 3

x = 3/2

c) x³ - 6x² + 11x - 6 = 0

x³ - x² - 5x² + 5x + 6x - 6 = 0

(x³ - x²) - (5x² - 5x) + (6x - 6) = 0

x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0

(x - 1)(x² - 5x + 6) = 0

(x - 1)(x² - 2x - 3x + 6) = 0

(x - 1)[(x² - 2x) - (3x - 6)] = 0

(x - 1)[x(x - 2) - 3(x - 2)] = 0

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

x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x - 3 = 0

x = 3

Vậy x = 1; x = 2; x = 3