Câu hỏi của Jimin

Question

tinh gia tri bieu thuc:

a,3x^4+5x^2y^2+2y^4+2y^2 biet rang x^2+y^2=1

b,x^3+xy^2-x^2y-y^3+3 biet x-y=0

Nguyen Quynh Huong · Accepted Answer

b, Ta co: $x^3+xy^2-x^2y-y^3+3$

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

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

= 3 ( vì x-y = 0)Câu trả lời: b, Ta co: $x^3+xy^2-x^2y-y^3+3$

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

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

= 3 ( vì x-y = 0)