Question

Cho x^2+y^2=1. C/m rằng b.thức 2(x^6+y^6)-3(x^4+y^4) ko phụ thuộc vào x,y.

Answer 1

=2(x²+y²)(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2(x⁴-x²y²+y⁴)-3x⁴-3y⁴

=2x⁴-2x²y²+2y⁴-3x⁴-3y⁴

=-x⁴-2x²y²-y⁴

=-(x⁴+2x²y²+y⁴)

=-(x²+y²)=-(1)²=-1

Answer 2

$2(x^6+y^6)-3(x^4+y^4)$

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

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

$=2x^4-2x^2y^2+2y^4-3x^4-3y^4$

$=-x^4-2x^2y^2-y^4$

$=-(x^4+2x^2y^2+y^4)$

$=-(x^2+y^2)$

$=-(1)^2$

$=-1$