设xy≠0,x^2+y^2=1,如果I1=(x^4+y^4)/(x^6+y^6),I2=(x^4+y^4)/√(x^6+
2个回答
∵x²+y²=1
且xy≠0
∴0
相关问题
4xy+1-4x^2-y^2= x^6-y^6-2x^3+1= x^4+64= x^3-4xy^2-2x^2y+8y^3
(1).设A=2X²-3Xy+y²+X-3y,B=4X²-6xy+2y²+4X-
若|x-1|+|y-3|=0,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+1/(x+6)(y+6).
若|x-1|+|y-3|=0,求1/xy+1/(x+2)(y+2)+1/(x+4)(y+4)+1/(x+6)(y+6).
先化简,再求值4x2y-[6xy-2﹙4xy-2﹚-x2y],其中x、y满足I2x-1I+﹙y+2﹚2=0
4x2y-[6xy-2(4xy-2)-x2y]+1,其中x=-[1/2],y=-1.
设A=2x²-3xy+y²+x-3y ,B=4x²-6xy+2y²+4x-y ,
4x²Y-(6X³+6XY²-2X²Y)+(2Y²X+6X³
1.设A=2x²-3xy+y²-x+2y,B=4x²-6xy+2y²-3x-y,
分解因式:(1):2x^2+xy-6y^2-x-16y-10 (2):x^2-4xy+4y^2+x-2y-6