a) \(3x^3-12x=0\)

=> \(3x\left(x^2-4\right)=0\)

=> \(\orbr{\begin{cases}3x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

b) \(x^2\left(x-3\right)+12-4x=0\)

=> \(x^2\left(x-3\right)+\left(-4x+12\right)=0\)

=> \(x^2\left(x-3\right)-4x+12=0\)

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

=> \(\left(x-3\right)\left(x^2-4\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)

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

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

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

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

d) \(x^2-4x-21=0\)

=> \(x^2+3x-7x-21=0\)

=> \(x\left(x+3\right)-7\left(x+3\right)=0\)

=> (x + 3)(x - 7) = 0 => x = -3 hoặc x = 7

e) 3x² - 7x - 10 = 0

=> 3x² + 3x - 10x - 10 = 0

=> 3x(x + 1) - 10(x + 1) = 0

=> (x + 1)(3x - 10) = 0

=> x = -1 hoặc x = 10/3

a) \(3x^3-12x=0\)

\(\Leftrightarrow3x\left(x^2-4\right)=0\)

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

\(\Rightarrow x\in\left\{-2;0;2\right\}\)

b) \(x^2\left(x-3\right)+12-4x=0\)

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

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

\(\Leftrightarrow x\in\left\{-2;2;3\right\}\)

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

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

\(\Leftrightarrow x\in\left\{-2;\frac{4}{5}\right\}\)

d) \(x^2-4x-21=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

e) \(3x^2-7x-10=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{10}{3}\end{cases}}\)

Ta có : 3x³ - 12x = 0

=> 3x(x² - 4) = 0

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

=> \(x\in\left\{0;2;-2\right\}\)

b) x²(x - 3) + 12 - 4x = 0

=> x²(x - 3) - 4(x - 3) = 0

=> (x² - 4)(x - 3) = 0

=> \(\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}\)

Vậy \(x\in\left\{-2;2;3\right\}\)

c) (3x - 1)² - (2x - 3)² = 0

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

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

=> \(\orbr{\begin{cases}x+2=0\\5x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0,8\end{cases}}\)

Vậy \(x\in\left\{-2;0,8\right\}\)

d) x² - 4x - 21 = 0

=> x² - 7x + 3x - 21 = 0

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

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

=> \(\orbr{\begin{cases}x+3=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=7\end{cases}}\)

Vậy \(x\in\left\{-3;7\right\}\)

e) 3x² - 7x - 10 = 0

=> 3x² + 3x - 10x - 10 = 0

=> 3x(x + 1) - 10(x + 1) = 0

=> (3x - 10)(x + 1) = 0

=> \(\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}\)

Vậy \(x\in\left\{\frac{10}{3};-1\right\}\)

5-x-16=40+x

-x-x=40+16-5

(-x).2=51

-x=51/2=>x=51/2

1.x=40

2.x=10

3.x=

15-x = 4x -5 

15 + 5 = 4x + x 

5x = 20

x= 4

x - 15 = 6 + 4x 

-15 - 6 = 4x -x 

3x = -21 

x= -7

c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)

=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12

=>-16x-1=-15x+12

=>-x=13

=>x=-13

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

a: \(x^3-4x^2-x+4=0\)

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

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

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

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

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

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

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

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

a, |3x - 6| + 4x + 3 = 12

|3x - 6| = 12 - 3 - 4x

|3x - 6| = 9 - 4x

TH1: 3x - 6 = 9 - 4x ⇔ 3x + 4x = 9 + 6 ⇔ 7x = 15 ⇔ x = 15/7 

TH2: 3x - 6 = 4x - 9 ⇔ 3x - 4x = -9 + 6 ⇔ -x = -3 ⇔ x = 3. 

b, |6x - 12| + 3x - 1 = 43

|6x - 12| = 43 + 1 - 3x

|6x - 12| = 44 - 3x

TH1: 6x - 12 = 44 - 3x ⇔ 6x + 3x = 44 + 12 ⇔ 9x = 56 ⇔ x = 6.

TH2: 6x - 12 = 3x - 44 ⇔ 6x - 3x = -44 + 12 ⇔ 3x = -32 ⇔ x = -32/3.

c, |8 - 4x| + 7x - 5 = 16 - 2x

|8 - 4x| = (16 + 5) + (-2x - 7x)

|8 - 4x| = 21 - 9x

TH1: 8 - 4x = 21 - 9x ⇔ -4x + 9x = 21 - 8 ⇔ 5x = 13 ⇔ x = 13/5

TH2: 8 - 4x = 9x - 21 ⇔ -4x - 9x = -21 - 8 ⇔ -15x = -29 ⇔ x = 29/15