Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

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

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)

i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)

g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)\)

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

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=6x^2\left(x+1\right)\)

g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)

câu f có ( x+1) là nhân tử chung

câu g đổi dấu - thành + thì (y-x) sẽ thành (x-y)

e: \(8\left(x+3y\right)-16x\left(x+3y\right)=8\left(x+3y\right)\left(1-2x\right)\)

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=6x^2\left(x+1\right)\)

Lời giải:
a. $5x^2-10xy=5x(x-2y)$

b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$

d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$

e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$

f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$

e) \(8\left(x+3y\right)-16x\left(x+3y\right)=\left(x+3y\right)\left(8-16x\right)=8\left(x+3y\right)\left(1-2x\right)\)

f) \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=\left(x+1\right)\left(4x^2+2x^2\right)=6x^2\left(x+1\right)\)

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

mn ơiiiiiiiiiiiii

 a) 3x2 – 7x + 2

\(=3x^2-6x-x+2\)

\(=\left(3x^2-6x\right)-\left(x-2\right)\)

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

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

 b) a(x2 + 1) – x(a2 + 1)

\(=ax^2+a-\left(a^2x+x\right)\)

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

.......?

a) Ta có: \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

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

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

b) Ta có: \(a\left(x^2+1\right)-x\left(a^2+1\right)\)

\(=x^2a+a-a^2x-x\)

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

\(=xa\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(xa-1\right)\)

c) Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

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

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24\)

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

\(=\left(x^2+7x\right)^2+16\left(x^2+7x\right)+6\left(x^2+7x\right)+96\)

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

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

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

d) Ta có: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+105+15\)

\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)^2+12\left(a^2+8a\right)+10\left(a^2+8a\right)+120\)

\(=\left(a^2+8a\right)\left(a^2+8a+12\right)+10\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+12\right)\left(a^2+8a+10\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

a) 3x . ( x-1 ) = 3x² - 3x 

b) x³- 2x²+x = x².( x-1 ) - x.( x-1 ) = (x-1).(x-1).x 

= (x-1)².x 

c) x²- 2xy-9z²+y²

= (x²-2xy+y² )-(3z)²

= (x-y)²-(3z)²

= ( x-y-3z).(x-y+3z)

thay vào ta có ( 6+4-90 ).(6+4+90 )=-80.100=-8000 

`a)5(x-y)-y(x-y)`

`=(x-y)(5-y)`

`b)x^2-6x-y^2+9`

`=(x^2-6x+9)-y^2`

`=(x-3)^2-y^2`

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