Question

biết x+y-2=0

tính N=\(x^3-2x^2-xy^2+2xy+2x+2y-2\)

Answer 1

Lời giải:

\(x+y-2=0\Leftrightarrow x+y=2\)

Suy ra:

\(N=x^3-2x^2-xy^2+2xy+2x+2y-2\)

\(=x^3-(x+y)x^2-xy^2+2xy+2x+2y-2\)

\(=-x^2y-xy^2+2xy+2(x+y)-2\)

\(=-xy(x+y)+2xy+2.2-2\)

\(=-2xy+2xy+4-2=2\)

Answer 2

\(N=x^3-2x^2-xy^2+2xy+2x+2y-2\)

\(N=\left(x^3-2x^2+x^2y\right)+\left(-x^2y+2xy-xy^2\right)+\left(2x+2y-4\right)+\left(4-2\right)\)

\(N=x^2\left(x-2+y\right)-xy\left(x-2+y\right)+2\left(x+y-2\right)+2\)

\(N=\left(x+y-2\right)\left(x^2-xy+2\right)+2=0.\left(x^2-xy+2\right)+2=2\)