已知x^2-2y^2=2,求(2x-y)(x+4y)-15的值.xy=0,忘打了
问题描述：
已知x^2-2y^2=2,求(2x-y)(x+4y)-15的值.
xy=0,忘打了
答
∵x^2-2Y^2=2 ∴x^2=2y^2+2
(2x-y)(x+4y)=2x(x+4y)-y(x+4y)=2x^+8xy-xy-4y^=2x^-4y^+7xy=2*2+7xy=4+7xy
4+7xy-15=7xy-11
答
x²-2y²=2
(2x-y)(x+4y)-15=2x²+8xy-xy-4y²-15=2(x²-2y²)+7xy-15=7xy-11
xy=0
所以=-11