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

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

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

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

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.

\(a,\left(3x+2\right)\left(2x-5\right)=\left(2x-5\right)\left(2x+5\right)\\ \Leftrightarrow\left(3x+2\right)\left(2x-5\right)-\left(2x-5\right)\left(2x+5\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(3x+2-2x-5\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\\ b,4x^2-8x=0\\ \Leftrightarrow4x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b: =>4x(x-2)=0

hay x=0 hoặc x=2

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

Bài 2: 

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

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

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^