Đáp án:
Giải thích các bước giải:
a) $\:x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)$
$=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3$
$=2x^2+x-x^3-2x^2+x^3-x+3$
$2x^2+x-x^3-2x^2+x^3-x+3$
$=3$
;
$\:x\left(3x^2-x+5\right)-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)$
$=3x^3-x^2+5x-\left(2x^3+3x-16\right)-x\left(x^2-x+2\right)$
$=3x^3-x^2+5x-2x^3-3x+16-x\left(x^2-x+2\right)$
$=3x^3-x^2+5x-2x^3-3x+16-x^3+x^2-2x$
$=16$
;
;
;
bài 2:
$=xy-xz+y\left(z-x\right)+z\left(x-y\right)$
$=xy-xz+yz-xy+z\left(x-y\right)$
$=xy-xz+yz-xy+xz-yz=0$
;
$=x\left(y+z-yz\right)-y\left(z+x-xz\right)+z\left(y-x\right)$
$=xy+xz-xyz-y\left(z+x-zx\right)+z\left(y-x\right)$
$=xy+xz-xyz-yz-xy+xyz+z\left(y-x\right)$
$=xy+xz-xyz-yz-xy+xyz+yz-xz$
$=0$
