Question

Cho x+y-2=0 . Tính giá trị BT sau  : x^4 + 2x^3y - 2x^3 +x^2y^2 - 2x^2y -x(x+y) + 2x+3

 

Answer 1

\(x^4+2x^3y-2x^3+x^2y^2-2x^2y-x(x+y)+2x+3\)

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

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

Do \(x+y-2=0\Rightarrow (x+y-2)(x^3+x^2y-x)=0\)

Answer 2

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

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

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

Do \(x+y-2=0\Rightarrow\left(x+y-2\right)\left(x^3+x^2y-x\right)=0\)

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