[(2x+2-2)]/(x+1),这一步没看懂,不是应该化成(x²+x-2)/x+1吗?
追答哦 对的,我看错了
x-(x+1)分之2 通分
=[x(x+1)-2]/(x+1)
=(x²+x-2)/(x+1)
=(x+2)(x-1)/(x+1)
(x²+2x)/(1+x)÷{x-2/(x+1)}-1
=x(x+2)/(x+1) ÷[(2x+2-2)]/(x+1) -1
=x(x+2)/(x+1)÷ (x+2)(x-1)/(x+1)-1
=x/(x-1)-1
=(x-x+1)/(x-1)
=1/(x-1)
=1/(√2+1-1)
=1/√2
=√2/2
1/(√2+1-1)这步错了吧,x的值不是(√2+1)吧!
x=根号2+1分之1
先化简,再求值:【(1+x)\/(x⊃2;+x+2)】÷【x-2+3\/(x+2)】,其中x=...
=【(1+x)\/(x²+x+2)】*【x+2\/(x+1)(x-1)】=(x+2)\/【(x-1)(x²+x+2)】=(1+2\/x)\/【(1-1\/x)(x²+x+2)】把x=1\/2带入,原式=(1+4)\/【(1-2)(1\/4+1\/2+2)】=5\/(-1*11\/4)=-5*4\/11=-20\/11 ...
...分之x]÷[(x²-1)分之(x+1)+1],其中x=根号2+1
=[x\/(x-1)²]÷[x\/(x-1)]=1\/(x-1)=1\/(√2+1-1)=1\/(√2)=(√2)\/2
先化简 再求值:(x+1) \/ x 除以{x-(1+ x的平方\/2x)},其中x=√2+1
=[(x+1)\/x]÷[2x²-1-x²)\/2x]=2\/(x-1)将x=√2+1带入得 2\/(x-1)=2\/(√2+1-1)=√2
(详细过程)先化简再求值(1+1\/x-2)÷x²-1\/2x-4,其中x=根号2-1
=(x-2+1)\/(x-2)÷(x-1)(x+1)\/2(x-2)=(x-1)\/(x-2)×2(x-2)\/(x-1)(x+1)=2\/(x+1)=2\/(√2-1+1)=2\/(√2)=(√2)\/2
...次方分之(1+x)的2次方÷(1-x分之2x-x)其中x=根号2
解析:已知x=根号2,那么:(1-x²)分之(1+x)² ÷ [(1-x)分之2x -x]=(1-x)分之(1+x) ÷ [(1-x)分之(2x+x²-x)]=(1-x)分之(1+x) ÷ [(1-x)分之(x²+x)]=(1-x)分之(1+x) × [(x²+x)分之(1-x)]=x分之1 =(根号2)分之...
先化简再求值x分之1+x除(x-(1+x平方)\/x平方),其中x=根号2加1
=1\/x+x\/[x-(1+x^2)\/x^2]=1\/x+x\/(x-1\/x^2-1)=(x-1\/x^2-1+x^2)\/(x^2-1\/x-x)=[(x-1)+(x+1\/x)(x-1\/x)] \/ [x(x-1)-1\/x]x值比较特殊, x=√2+1, 所以x的倒数1\/x=√2-1 这样代入的时候就是比较容易求值了 =[√2+2√2.2] \/ [(√2+1).√2-...
...2x+1)分之x+(1-x)分之1】÷(x²-1)分之2?
\/(x-1)×(x+1)(x-1)\/2 =(x+1)\/2 当x=-2时 原式=(-2+1)\/2=-1\/2,5,【先化简,再求值】【(x²-2x+1)分之x+(1-x)分之1】÷(x²-1)分之2 【先化简,再求值】【(x²-2x+1)分之x+(1-x)分之1】÷(x²-1)分之2 其中x=-2 ...
先化简,再求值:(1-x+1分之1)÷x2-1分之x,其中x=-2分之3
原式=(x+1-1)\/(x+1)*(x²-1)\/x =x\/(x+1)*(x+1)(x-1)\/x =x-1 =-2分之3-1 =-2分之5
先化简,再求值:(x一1分之2x一x+1分之x)÷X²一1分之x,其中x=2
(x一1分之2x一x+1分之x)÷X²一1分之x =[2x\/(x-1) -x\/(x+1)]*(x2-1)\/x =2*(x+1)-(x-1)=x+3 x=2,原式=2+3=5
...1+x-1分之1)÷x²-2x+1分之x²-x,其中x=根号2-1
1 2016-05-29 先化简(x-1分之x+1+1)÷x^2-2X+1分之x^2+... 4 2017-02-13 先化简,再求值 x+1分之1 2017-12-24 先化简x²-2x+1分之x²+x ÷(x... 2 2014-11-08 先化简再求值:(x的平方-1分之x+1 + x-1分之x)÷...更多...