初二的数学题~~~急,速度~~~积分高1.已知ab＝1,求1/(a²-1)+1／(b²-1)的值2.已知1/x+1/y=-1/(x+y),求y/x+x/y的值3.已知(x+3)/(x+2)=1/(√3+√2+1),求(x-3)/(2x-4)÷[5/(x-2)-x-2]的值

答案:1.a=1/b,所以原式=1/(1/b^2-1)+1/(b^2-1)=-b^2(b^2-1)+1/(b^2-1)=-1
2.两边乘以xy,得x+y=-xy/(x+y),即(x+y)^2/xy=-1,即x^2+y^2/xy=-3,所以y/x+x/y=x^2+y^2/xy=-3 我是读初二的！好好学习啊！

1.已知ab＝1，求1/(a²-1)+1／(b²-1)的值
a=1/b,所以原式=1/(1/b^2-1)+1/(b^2-1)
=-b^2(b^2-1)+1/(b^2-1)
=-1
2.已知1/x+1/y=-1/(x+y),求y/x+x/y的值
1/x+1/y=1/(x+y)
(x+y)/(xy)=1/(x+y)
(x+y)^2=xy
y/x+x/y
=(y^2+x^2)/(xy)
=(x+y)^2/(xy)-2xy/(xy)
=1-2
=-1
3.已知(x+3)/(x+2)=1/(√3+√2+1),求(x-3)/(2x-4)÷[5/(x-2)-x-2]的值
(X-3)\(2X-4)÷[5\(x-2）－X－2]通分化简
=(x-3)/(2x-4)*(x-2)/(5-(x^2-4))=(x-3)/((3-x)(6+2x))=-1/(6+2x).
又由已知（X＋3）\(X+2)=1\(根3＋根2＋1)得
(x+2)/(x+3)=根3＋根2＋1=(x+3-1)/(x+3)=1-1/(x+3)。所以。。
-1/(x+3)=根3＋根2
(X-3)\(2X-4)÷[5\(x-2）－X－2]=-1/(6+2x)=(根3＋根2)/2

我是读初二的!1.a=1/b,原式=1/(1/b^2-1)+1/(b^2-1)=-b^2(b^2-1)+1/(b^2-1)=-1
2.两边乘以xy,得x+y=-xy/(x+y),即(x+y)^2/xy=-1,即x^2+y^2/xy=-3,所以y/x+x/y=x^2+y^2/xy=-3

1.a=1/b,所以原式=1/(1/b^2-1)+1/(b^2-1)=-b^2(b^2-1)+1/(b^2-1)=-1
2.两边乘以xy,得x+y=-xy/(x+y),即(x+y)^2/xy=-1,即x^2+y^2/xy=-3,所以y/x+x/y=x^2+y^2/xy=-3