Bài 1:
a.
$3x(x-y)-3y(y-x)=3x(x-y)+3y(x-y)=(x-y)(3x+3y)=3(x-y)(x+y)$
b.
$5x+5y+x^2+2xy+y^2=(5x+5y)+(x^2+2xy+y^2)$
$=5(x+y)+(x+y)^2=(x+y)(5+x+y)$
c.
$x^3-4x=x(x^2-4)=x(x^2-2^2)=x(x-2)(x+2)$
d.
$x^2-16+y^2+2xy=(x^2+2xy+y^2)-16$
$=(x+y)^2-4^2=(x+y-4)(x+y+4)$
e.
$2x^2-5x-7=(2x^2+2x)-(7x+7)=2x(x+1)-7(x+1)=(x+1)(2x-7)$
f.
$x^3+2x^2y+xy^2-25x=x(x^2+2xy+y^2-25)$
$=x[(x^2+2xy+y^2)-5^2]=x[(x+y)^2-5^2]=x(x+y-5)(x+y+5)$
Bài `2`
`x^3 -9x=0`
`<=>x(x^2-9)=0`
`<=>x(x-3)(x+3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
`-------------`
`6x (x-3)+ 3-x=0`
`<=>6x (x-3)+(3-x)=0`
`<=>(x-3)(6x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\6x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{6}\end{matrix}\right.\)
`-------------`
`3x(x-1) -x+1=0`
`<=> 3x(x-1) -(x-1)=0`
`<=>(x-1)(3x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
giúp mình bài 3 vs mn ơi ko phải thi chỉ là đề ôn thôi ạ
