Question

Cho x² + y² =1. Tính:

1, A= 2(x⁶ + y⁶ ) - 3(x⁴ + y⁴)

2, B = 2x⁴ - y⁴ + x²y² + 3y²

Answer 1

1,Ta có: \(A=2\left[\left(x^2\right)^3+\left(y^2\right)^3\right]-3x^4-3y^4\)

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

Thay \(x^2+y^2=1,\) ta có:

\(A=2.1\left(x^4-x^2y^2+y^4\right)-3x^4-3y^4\)

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

\(=-\left(x^4+2x^2y^2+y^4\right)=-\left(x^2+y^2\right)^2=-1\)

2,Ta có: \(B=\left(x^2-y^2\right)\left(x^2+y^2\right)+\left(x^4+x^2y^2\right)+3y^2\)

\(=\left(x^2-y^2\right).1+x^2\left(x^2+y^2\right)+3y^2\)

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