Giải thích các bước giải:

5.Ta có:

$3x(2x+1)-20=6x(x-2)+2$ 

$\to 6x^2+3x-20=6x^2-12x+2$

$\to 15x=22$

$\to x=\dfrac{22}{15}$

6.Ta có:

$3x^3-x^2-x(-2x^2+3x-1)$

$=3x^3-x^2+2x^3-3x^2+x$

$=3x^3+2x^3-x^2-3x^2+x$

$=5x^3-4x^2+x$

7.Ta có:

$x^3-5x^2+2x(-x^2+3x-5)$

$=x^3-5x^2-2x^3+6x^2-10x$

$=x^3-2x^3-5x^2+6x^2-10x$

$=-x^3+x^2-10x$

8.Ta có:

$-7x^4+5x^3-x^2(-2x^2+3x-1)$

$=-7x^4+5x^3+2x^4-3x^3+x^2$

$=-7x^4+2x^4+5x^3-3x^3+x^2$

$=-5x^4+2x^3+x^2$

9.Ta có:

$x^4+3x^3+2x(-x^2-x+1)$

$=x^4+3x^3-2x^3-2x^2+2x$

$=x^4+x^3-2x^2+2x$

10.Ta có:

$3x^4-4x^3+2x(x^3-2x^2+7x)$

$=3x^4-4x^3+2x^4-4x^3+14x^2$

$=3x^4+2x^4-4x^3-4x^3+14x^2$

$=5x^4-8x^3+14x^2$