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

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

\(x^2-9-\left(x-3\right)\left(5-2x\right)=\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5-2x\right)=\left(x-3\right)\left(x+3-5+2x\right)=\left(x-3\right)\left(3x-2\right)\)

\(x^2-9-\left(x-3\right)\left(5-2x\right)\\ =\left(x^2-9\right)-\left(x-3\right)\left(5-2x\right)\\ =\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5-2x\right)\\ =\left(x-3\right)\left(x+3-5+2x\right)\\ =\left(x-3\right)\left(3x-2\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=y\)

\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(y+1\right)\left(y-1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\\ =\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+16\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\\ =\left(x^2+7x+11\right)^2-1-24\\ =\left(x^2+7x+11\right)^2-25\\ =\left(x^2+7x+16\right)\left(x^2+7x+6\right)\\ =\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

a: =(x-y)(5-y)

b: \(=x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

\(a,5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\\ b,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

Ta có \(f\left(x\right)=\left(x+1\right)\left(x-3\right)\left(x+5\right)\left(x+9\right)+256\)

\(f\left(x\right)=\left(x+1\right)\left(x+5\right)\left(x-3\right)\left(x+9\right)+256\)

\(f\left(x\right)=\left(x^2+6x+5\right)\left(x^2+6x-27\right)+256\)

\(f\left(x\right)=\left[\left(x^2+6x-11\right)^2-256\right]+256\)

\(f\left(x\right)=\left(x^2+6x-11\right)^2\)

`a)3x-3a+yx-ya`

`=3(x-a)+y(x-a)`

`=(x-a)(y+3)`

`b)x^2-9-4(x+3)`

`=(x-3)(x+3)-4(x+3)`

`=(x+3)(x-3-4)`

`=(x+3)(x-7)`

a)  \(=3x+yx-\left(3a+ya\right)\) \(=x\left(3+y\right)-a\left(3+y\right)\) \(=\left(3+y\right)\left(x-a\right)\)

b) \(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)  \(=\left(x+3\right)\left(x-3-4\right)\) \(=\left(x+3\right)\left(x-7\right)\)