Đáp án:
Giải thích các bước giải:
$1.(x-1)^3-x(x-2)^2+1$
$=x^3-3x^2+3x-1-x(x^2-4x+4)+1$
$=x^3-3x^2+3x-1-x^3+4x^2-4x+1$
$=(x^3-x^3)-(3x^2-4x^2)+(3x-4x)-(1-1)$
$=x^2-x$
$2.2x(3x+2)-3x(2x+3)$
$=6x^2+4x-6x^2-9x$
$=-5x$
$3.(x+2)^3+(x-3)^2-x^2(x+5)$
$=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2$
$=2x^2+6x+17$
$4.(2x+3)(x-5)+2x(3-x)+x-10$
$=2x^2-10x+3x-15+6x-2x^2+x-10$
$=(2x^2-2x^2)-(10x-6x-x-3x)-(15+10)$
$=-25$
Chúc em học tốt.