要初二数学计算题！！越多越好！！

来源：百度知道 编辑：UC知道 时间：2024/06/23 20:44:23
不要网址的，要现题。
要例如：2－X÷(X－1) ÷『X＋1－3÷(X－1)』或X平方＋9Y平方－8X＋6Y=0求1/X－1/Y的值。这种题的

(x^3-1/x^3+2x^2+2x+1)+(x^3+1/x^3-2x^2+2x-1)-2(x^2+1/x^2-1)
解(x^3-1)/(x^3+2x^2+2x+1)+(x^3+1)/(x^3-2x^2+2x-1)-[2(x^2+1)]/(x^2-1)
=(x^3-1)/[x^3+1+2x(x+1)]+(x^3+1)/[x^3-1-2x(x-1)]-[2(x^2+1)]/(x^2-1)
=(x^3-1)/[(x+1)(x^2+1+x)+2x(x+1)]+(x^3+1)/[(x-1)(x^2+1-x)-2x(x-1)]
-[2(x^2+1)]/(x^2-1)
=(x^3-1)/(x+1)[(x^2+1-x)+2x]+(x^3+1)/(x-1)[(x^2+1+x)-2x]
-[2(x^2+1)]/(x^2-1)
=(x^3-1)/[(x+1)(x^2+1+x)]+(x^3+1)/[(x-1)(x^2+1-x)]
-[2(x^2+1)]/(x^2-1)
=[(x-1)(x^2+1+x)]/[(x+1)(x^2+1+x)]+[(x+1)(x^2+1-x)]/[(x-1)(x^2+1-x)]
-[2(x^2+1)]/(x^2-1)
=(x-1)/(x+1)+(x+1)/(x-1)
-[2(x^2+1)]/(x^2-1)
=[(x-1)^2+(x+1)^2-2(x^2+1)]/(x^2-1)
=[(x^2-2x+1)+(x^2+2x+1)-2x^2-2)]/(x^2-1)
=[(-2x+1)+(2x+1)-2]/(x^2-1)
=(1+1-2)/(x^2-1)

x^4+16y^28x^2y 二1/2x^2+2xy+2y^2
解x^4+16y^2+8x^2y
=(x^2)^2+2*x^2*(4y)+(4y)^2
=(x+4y)^2

1/2x^2+2xy+2y^2
=1/2(x^2+4xy+4y^2)
=1/2(x+2y)^2

a/(根号ab+b)+b/(根号