已知X+Y+2=0,求8(X^2+Y^2)-(X^2-Y^2)^2的值.

熱心網友

解：已知X+Y+2=0，得X+Y=-2化簡原式：原式=8(X^2+Y^2)-4(X-Y)^2=4(X^2+2xy+Y^2)=4(X+Y)^2將X+Y=-2代入原式，得：原式=4*（-2）^2=4*4=16（最標準的格式、過程和答案）

熱心網友

8(X^2+Y^2)-(X^2-Y^2)^2==8x^2+8y^2-[(x+y)^2][(x-y)^2]=8x^2+8y^2-4(x-y)^2 (因：x+y=-2)=8x^2+8y^2-4x^2-4y^2+8xy=4(x^2+2xy+y^2)=4(x+y)^2=4×4=16

熱心網友

解：已知X+Y+2=0，得X+Y=-2所以8(X^2+Y^2)-(X^2-Y^2)^2=8(X^2+Y^2)-(X+Y)^2)*(X-Y)^2 =8(X^2+Y^2)-4(X-Y)^2 =4(X^2+2xy+Y^2) =4(X+Y)^2 =16

熱心網友

x+y=-2 (x+y)^2=4 8(X^2+Y^2)-(X^2-Y^2)^2=8(X^2+Y^2)-(X+Y)^2(x-y)^2=8(x^2+y^2)-4(x-y)^2=4(x^2+y^2+2xy)=4(x+y)^2=4*4=16