(3x/x+1-x/x-1)+x-2/x的平方-1
=(3x(x-1)-x(x+1))/(x+1)(x-1)+x-2/x^2-1
=(3x^2-3x-x^2-x+x-2)/(x^2-1)
=(2x^2-3x-2)/(x^2-1)
=(2(√3/2)^2-3(√3/2)-2)/((√3/2)^2-1)
=(3/2-3√3/2-2)/(3/4-1)
=(-1/2-3√3/2)/(-1/4)
=2+6√3
3x-3/x平方-1 除3x/x+1 减1/x-1
=3(x-1)/[(x-1)(x+1)]*(x+1)/3x-1/(x-1)
=1/x-1/(x-1)
=1/2-1
=-1/2
解答题
解
x/(x-1)-3/(x-1)(x+2)-1
=x(x+2)/(x-1)(x+2)-3/(x-1)(x+2)-1
=(x²+2x-3)/(x-1)(x+2)-1
=(x+3)(x-1)/(x-1)(x+2)-1
=(x+3)/(x+2)-(x+2)/(x+2)
=(x+3-x-2)/(x+2)
=1/(x+2)
x+1分之x平方-x乘x平方-2x+1分之x平方-1
=x(x-1)/(x+1)乘以(x+1)(x-1)/(x-1)²
=x
当x=根号2时
原式=根号2
原式=x^2-3x+2-3x^2-9x+2x^2+2x-4
=-10x-2
当x=1/3时
原式等于-16/3
(3x+4/x的平方-1-2/x-1)÷x+2/x的平方-2x+1,
=[(3x+4-2(x+1))/(x^2-1)] *(x^2-2x+1)/(x+2)
=[(3x+4-2x-2)/(x-1)(x+1)] * (x-1)^2/(x+2)
=[(x+2)/(x-1)(x+1)] * (x-1)^2/(x+2)
=(x-1)/(x+1)
其中x是方程8x-5=3x的解
则有x=1
上式=(1-1)/(1+1)=0
应该先求定义域,x-1>=0,1-x>=0同时成立,则x=1,再代回得y=-1,又将其代入下个式子:
好像的-5/2,(自己算一下嘛)
((x-1)/x-(x-2)/(x+1))÷(2x平方 –x)/(x平方 +2x+1)
=(x²-1-x²+2x)/[x(x+1)]×(x+1)²/x(2x-1)
=(2x-1)(x+1)²/[x²(x+1)(2x-1)]
=(x+1)/x²
因为
x²-x-1=0
即
x²=x+1
所以
原式=x²/x²=1
由x²+x-6=0
解得x1=2,x2=-3,
(x+1-3/(x-1))÷(x平方-4x+4)/(x-1)
=(X-2)(X+2)/[(x+1)(x-1)]*(x-1)/(x-2)²
=(X+2)/[(X+1)(X-2)]
当X=2时,原式无意义;
当X=-3时,
原式=(-3+2)/[(-3+1)(-3-2)]
=-1/10
有疑问,请追问;若满意,请采纳,谢谢!