Đáp án:
$1)
a) (x-2)(4x-8+y)$
$b)(x-y).\left [x^3-2x^2y+xy^2-xy \right ]\\
c)
(x+3)(2ax^2+6a)\\
d) (a+b)(3x^2+x+5)\\
2)
D=x^2(x+2y)(x^3-xy+y^2)\\
3)
a) {\left[\begin{aligned}x=4\\x=\dfrac{7}{2}\end{aligned}\right.}\\
b)x=2$
Giải thích các bước giải:
$1)
a) 4(2-x)^2+xy-2y=4(x-2)^2+y(x-2)=(x-2)(4x-8+y)\\
b) x(x-y)^3-y(y-x)^2-y^2(x-y)=x(x-y)^3-y(x-y)^2-y^2(x-y)=(x-y)\left [x(x-y)^2-y(x-y)-y^2 \right ]\\
=(x-y)\left [x(x^2-2xy+y^2-xy+y^2-y^2 \right ]\\
=(x-y)\left [x^3-2x^2y+xy^2-xy \right ]\\
c)
2ax^3+6ax^2+6ax+18a=2ax^2(x+3)+6a(x+3)=(x+3)(2ax^2+6a)\\
d) 3ax^2+3bx^2+ax+bx+5a+5b=3x^2(a+b)+x(a+b)+5(a+b)=(a+b)(3x^2+x+5)\\
2)
D=x^5(x+2y)-x^3y(x+2y)+x^2y^2(x+2y)=x^2(x+2y)(x^3-xy+y^2)\\
3)
a) 4-x=2(x-4)^2\\
\Leftrightarrow 2(x-4)^2+x-4=0\\
\Leftrightarrow (x-4)(2x-8+1)=0\\
\Leftrightarrow (x-4)(2x-7)=0\\
\Leftrightarrow {\left[\begin{aligned}x-4=0\\2x-7=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=4\\x=\dfrac{7}{2}\end{aligned}\right.}\\
b)
2-x=2(x-2)^3\\
\Leftrightarrow 2(x-2)^3+x-2=0\\
\Leftrightarrow (x-2)[2(x-2)^2+1]=0\\
\Leftrightarrow {\left[\begin{aligned}x-2=0\\(x-2)^2=\dfrac{-1}{2} (VL\end{aligned}\right.}\\
\Leftrightarrow x=2$
