1.已知4x^2-3y^2=7,3x^2+3y^2=19,求代数式14x^2-2y^2的值.

1.
4x^2-3y^2=7 ,(1)
3x^2+3y^2=19,(2)
(1)+(2)
7x^2=26
x^2=26/7 (3)
将(3)代入(1)中
4×(26/7)-3y^2＝7
-3y^2=7-106/7
y^2=19/7
因此 14x^2-2y^2=2×26-2×(19/7)=52-38/7=326/7
2.
xy+x=-1,(1)
xy-y=-2,(2)
(1)-(2)
x+y=1 (3)
由(3)可得
x=1-y(4)
y=1-x(5)
-x-[2y-2(xy+x)^2+3x]+2[x+xy-y)]^2
将xy+x=-1,xy-y=-2代入,则
原式=-x-[2y-2×(-1)^2+3x]+2[x-2]^2
=-x-2y+2-3x+2×(2-x)^2
=-4x-2y+2+2×(1+1-x)^2
将(5)代入,则
=-4x-2y+2+2×(1+y)^2
=-4x-2y+2+2+4y+2y^2
=4y-4x-2y+2y^2+4
=2y+2y^2+4×(1-x)
将(5)代入,则
=2y+2y^2+4y
=6y+2y^2