a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

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

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

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

=>x+2=0

=>x=-2

huhu , ai rảnh thì làm giúp mk vs ạ !!!

a)  \(x^3+8x^2+17x+10\)

\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+7x+10\right)\)

\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

b)  \(=x^4-2x^3-12x^2+12x+36\)

\(=x^2\left(x^2-2x-6\right)-2\left(x^2-2x-6\right)\)

\(=\left(x^2-2\right)\left(x^2-2x-6\right)\)

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

1. 5x² - 4x

= x(5x - 4)

2. 8x²(x - 3y) - 12x(x - 3y)

= (8x² - 12x)(x - 3y)

= 4x(2x - 3)(x - 3y)

3. 3(x - y) - 5x(y - x)

= 3(x - y) + 5x(x - y)

= (3 + 5x)(x - y)

\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)

\(=\left(x^2-x-6\right)^2-x^2=\left(x^2-6\right)\left(x^2-2x-6\right)\)

Bài 4:

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

\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)

\(x^2-12x+36=\left(x-6\right)^2\)

a.\(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

b.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)

c.\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

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

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

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

\(\Leftrightarrow x=-2\)

\(4x^2-12x+5=4x^2-10x-2x+5=2x\left(2x-5\right)-\left(2x-5\right)=\left(2x-1\right)\left(2x-5\right)\)

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

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

\(4x^2-12x+5=4\left(x^2-3x+\frac{5}{4}\right)=4\left(x^2-3x+\frac{9}{4}-\frac{9}{4}+\frac{5}{4}\right)=4\left(x-\frac{3}{2}\right)^2-2=\left(2x-1\right)\left(2x-5\right)\)