Giải thích các bước giải:
b.Ta có:
$B=x^3-2x^2-xy^2+2xy+2y+2x-2$
$\to B=(x^3-xy^2)-2x^2+2xy+2(y+x)-2$
$\to B=x(x^2-y^2)-2x^2+2xy+2(y+x)-2$
$\to B=x(x-y)(x+y)-2x^2+2xy+2(y+x)-2$
$\to B=x(x-y)\cdot 2-2x^2+2xy+2\cdot 2-2$ vì $x+y-2=0\to x+y=2$
$\to B=2x^2-2xy-2x^2+2xy+4-2$
$\to B=(2x^2-2x^2)+(2xy-2xy)+(4-2)$
$\to B=0+0+2$
$\to B=2$
c.Ta có: $x+y=2\to y=2-x$
$\to C=x^4-2x^3(2-x)-2x^3+x^2(2-x)^2-2x^2(2-x)-x(x+2-x)+2x+3$
$\to C=x^4-4x^3+2x^4-2x^3+4x^2-4x^3+x^4-4x^2+2x^3-x\cdot \:2+2x+3$
$\to C=x^4+2x^4+x^4-4x^3-2x^3-4x^3+2x^3+4x^2-4x^2-2x+2x+3$
$\to C=4x^4-8x^3+3$
