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:

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

6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)

8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2  = -(y-x^2-x) (y+x^2+x)

10, 4x mũ 2 ( x + y ) -x - y  = (2x-1) (2x+1) (y+x)

Bài 2: 

c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)

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

a: \(16x^3+0,25yz^3\)

\(=0,25\cdot x^3\cdot64+0,25\cdot yz^3\)

\(=0,25\left(64x^3+yz^3\right)\)

b: \(x^4-4x^3+4x^2\)

\(=x^2\cdot x^2-x^2\cdot4x+x^2\cdot4\)

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

c: \(x^3+x^2y-xy^2-y^3\)

\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)\)

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

\(=\left(x-y\right)\cdot\left(x+y\right)^2\)

d: \(x^3+x^2+x+1\)

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

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

e: \(x^4-x^2+2x-1\)

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

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

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

f: \(2x^2-18\)

\(=2\cdot x^2-2\cdot9\)

\(=2\left(x^2-9\right)=2\left(x-3\right)\left(x+3\right)\)

g: \(x^2+8x+7\)

\(=x^2+x+7x+7\)

\(=x\left(x+1\right)+7\cdot\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)

h: \(x^4y^4+4\)

\(=x^4y^4+4x^2y^2+4-4x^2y^2\)

\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2y^2+2-2xy\right)\left(x^2y^2+2+2xy\right)\)

i: \(x^4+4y^4\)

\(=x^4+4x^2y^2+4y^4-4x^2y^2\)

\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)

k: \(x^2-2x-15\)

\(=x^2-5x+3x-15\)

\(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)