3x³ - 48x = 0

=> 3x( x² - 16) = 0

=> x = 0 hoặc x² -16 = 0

x² - 16 = 0 => x² = 16 => x = 4 hoặc x =-4

\(\text{1) }3x^3-48x=0\\ \Leftrightarrow x\left(3x^2-48\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2-48=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2=48\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\\ \text{Vậy }x=0\text{ hoặc }x=\pm4\)

\(\text{2) }x^3+x^2-4x=4\\ \Leftrightarrow x^3+x^2-4x-4=0\\ \Leftrightarrow\left(x^3+x^2\right)-\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\\ \Leftrightarrow\left(x^2-4\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\\ \text{Vậy }x=2\text{ hoặc }x=-2\text{ hoặc }x=1\)

1) \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

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

\(\Leftrightarrow3x\left(x-4\right)\left(x+4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Vậy x=0 ; x=4 ; x=-4

b) \(x^3+x^2-4x=4\)

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

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

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

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

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

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

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

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

Vậy x=-1 ; x=2 ; x=-2

a) Ta có: \(x^3+3x^2+3x+2=0\)

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

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

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

mà \(x^2+x+1\ne0\forall x\)

nên x+2=0

hay x=-2

Vậy: x=-2

b) Ta có: \(x^3-12x^2+48x-72=0\)

\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)

\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)

mà \(x^2-6x+12\ne0\forall x\)

nên x-6=0

hay x=6

Vậy: x=6

a, Thắc mắc đề cóa sai khong .

( đáp án vẫn có nhưng là số vô tỉ nên nghe lạ á )

b, Ta có : \(x^3-12x^2+48x-72=0\)

=> \(x^3-3.x^2.4+3.x.4^2-64-8=0\)

=> \(\left(x-4\right)^3-8=0\)

=> \(\sqrt[3]{\left(x-4\right)^3}=\sqrt[3]{8}=2\)

=> \(x=6\)

Vậy ....

\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\frac{48x^3}{3x^5}}=\sqrt{\frac{16}{x^2}}=\frac{4}{x}\)

a, 2x(x-5) - x ( 3 + 2x ) = 26

=> 2x^2 - 10x - 3x - 2x ^ 2 = 26 

=> - 13 x = 26 

=> x = -2

a, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Vậy x = -2

b, \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=-4\end{cases}}\)

Vậy x = 0 hoặc x = 4 hoặc x = -4 

a) 2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

-13x = 26

=> x = 2

b) 3x³ - 48x = x(3x² - 48) = 0

=> \(\orbr{\begin{cases}x=0\\3x^2-48=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x^2=48\Rightarrow x^2=16\end{cases}}}\Rightarrow\orbr{\begin{cases}x=0\\x=4;-4\end{cases}}\)

Vậy x = {0 ; 4 ; -4}

a) \(x^2-4=0\)

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

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

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

b) \(x\left(x+5\right)=9x\)

\(\Rightarrow x^2+5x-9x=0\)

\(\Rightarrow x^2-4x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

c) \(3x^3-48x=0\)

\(\Rightarrow3x\left(x^2-16\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\Rightarrow\left(x-4\right)\left(x+4\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

d) \(x^4+x^2-20=0\)

\(\Rightarrow\left(x^2\right)^2+x^2-20=0\)

Đặt x² = a

\(\Rightarrow a^2+a-20=0\)

\(\Rightarrow a^2+5a-4a-20=0\)

\(\Rightarrow a\left(a+5\right)-4\left(a+5\right)=0\)

\(\Rightarrow\left(a-4\right)\left(a+5\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2=4\Rightarrow x=\pm2\\x^2=-5\Rightarrow x\in\varnothing\end{matrix}\right.\)

d) x⁴ + x² - 20 = 0

\(\Rightarrow\) x⁴ + x² = 20

\(\Rightarrow\) x⁴ + x² = 2⁴ + 2²

\(\Rightarrow\) x = 2

a) x² - 4 = 0

\(\Rightarrow\) x² = 4

\(\Rightarrow\) x = 2 hoặc -2

\(3x^3-48x=8\)

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

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

\(\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)

\(x^2-2x=24\)

\(x^2-2x-24=0\)

\(x^2-6x+4x-24=0\)

\(x\left(x-6\right)+4\left(x-6\right)=0\)

\(\left(x+4\right)\left(x-6\right)=0\)

\(\left[\begin{array}{nghiempt}x+4=0\\x-6=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=-4\\x=6\end{array}\right.\)

a)\(3x^3-48x=0\\ \Rightarrow3x\left(x^2-16\right)=0\\ \Rightarrow3x\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}3x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Vậy x\(\in\){0;\(\pm\)4}

b)\(x^2-2x=24\\ \Rightarrow x^2-2x-24=0\\ \Rightarrow x^2-6x+4x-24=0\\ \Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\\ \Rightarrow\left(x-6\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy x\(\in\){6;\(-\)4}

a) 3x³ - 48x = 0

⇔3x.(x² - 16) = 0

⇔3x.(x² - 4²) = 0

⇔3x.(x-4).(x+4) = 0

⇔\(\left[{}\begin{matrix}3x=0\\x-4=0\\x+4=0\end{matrix}\right.\)\(\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Còn câu b) bạn tự làm nha.

\(1,x^3-3x^2=0\)

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

\(\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=3\left(TM\right)\end{cases}}}\)

\(2,3x^3-48x=0\)

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

\(\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x^2=16\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=\pm4\left(TM\right)\end{cases}}}}\)

\(3,5x\left(x-1\right)=x-1\)

\(5x^2-5x=x-1\)

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

\(5x^2-5x-x+1=0\)

\(5x\left(x-1\right)-\left(x-1\right)=0\)

\(\left(5x-1\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\orbr{\begin{cases}x=\frac{1}{5}\left(TM\right)\\x=1\left(TM\right)\end{cases}}}\)

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

\(2x+10-x^2-5x=0\)

\(-x^2-3x+10=0\)

\(-x^2-5x+2x+10=0\)

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

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

\(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}\orbr{\begin{cases}x=-5\left(TM\right)\\x=2\left(TM\right)\end{cases}}}\)

\(5,2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x-26=0\)

\(-13\left(x+2\right)=0\)

\(x=-2\left(TM\right)\)

Trả lời:

1, \(x^3-3x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

Vậy x = 0; x = 3 là nghiệm của pt.

2, \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}}\)

Vậy x = 0; x = 4; x = - 4 là nghiệm của pt.

3, \(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = 1; x = 1/5 là nghiệm của pt.

4, \(2\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

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

Vậy x = - 5; x = 2 là nghiệm của pt.

5, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Vậy x = - 2 là nghiệm của pt.