x²( x + 1 ) + 2x( x + 1 ) = 0 <=> x( x + 1 )( x + 2 ) = 0 <=> x = 0 hoặc x = -1 hoặc x = -2

x( 3x - 1 ) - 5( 1 - 3x ) = 0 <=> x( 3x - 1 ) + 5( 3x - 1 ) = 0 <=> ( 3x - 1 )( x + 5 ) = 0 <=> x = 1/3 hoặc x = -5

Trả lời:

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

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

\(\Leftrightarrow x=0;x=-1;x=-2\)

Vậy x = 0; x = - 1; x = - 2 là nghiệm của pt.

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-5\end{cases}}}\)

Vậy x = 1/3; x = - 5 là nghiệm của pt.

1,X=-1 hoặc 3

2,Tìm x sao cho (x+3) và (3x-2) ko bằng 0

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

a,\(\left(x-1\right)^2-\left(2x\right)^2=0< =>\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(< =>\left(-x-1\right)\left(3x-1\right)=0< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\left(3x-5\right)^2-x\left(3x-5\right)=0< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow x=-1;x=\frac{1}{3}\)

b, \(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x-5-x\right)=0\Leftrightarrow\left(3x-5\right)\left(2x-5\right)=0\Leftrightarrow x=\frac{5}{3};x=\frac{5}{2}\)

b: \(3x^2-2x-1=0\)

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

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

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

a: Bạn ghi lại đề đi bạn

a)  \(x^3+3x^2+3x+2=0\)

<=>  \(x^3+x^2+x+2x^2+2x+2=0\)

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

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

tự làm

b) \(x^4-2x^3+2x-1=0\)

<=>  \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)

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

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

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

tự làm

c)   \(x^4-3x^3-6x^2+8x=0\)

<=>   \(x\left(x^3-3x^2-6x+8\right)=0\)

<=>  \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)

<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)

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

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

tự làm

a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

Bài 2: 

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

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