(x-x分之1)除以x方分之x方-2x 1
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f(1/x)=1-x^2+1/x^2=-f(x)直接代入即可再问:能写的再具体点吗我不太明白谢谢再答:哦我把题目看错了你的题目是对的吗再问:嗯题目是对的加我Q1047763981这样说方便再答:我没有
原式=(x²+x-2x)/(x+1)÷[(x-1)²/(x+1)(x-1)]=(x²-x)/(x+1)÷[(x-1)/(x+1)]=x(x-1)/(x+1)*(x+1)/
x^2-x-6分解因式得(x+2)(x-3)x^2-4分解因式得(x-2)(x+2)故(x^2-x-6分之x^2-4+x-3分之x+2)=(x-2)/(x-3)+(x+2)/(x-3)=2x/(x-3
自己想
原式=[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x+3)(x+5)]\(x^2+7x+13)=[(1分之x+2)-(1分之x+3)+(1分之x+
把X方—3X+1=0两边除以X得X-3+x分之一=0,x+x分之一=3,变成x方分之一+2+x方=3,变成x方分之一+x方=1,
我操,你这样鬼能看懂啊?分子是啥分母是啥,那些加、除都在什么位置上啊?
=[(x+1)(x-1)/(x-1)²-(x-1)/(x+1)]×(x-1)/x=[(x+1)²-(x-1)²]/(x+1)(x-1)×(x-1)/x=4x/(x+1)(
把每个式子因式分解x的方+5x+4=(x+1)(x+4)x方-5x+6=(x-2)(x-3)………………
(x+y)/(x^4-y^4)÷1/(x^2+y^2)=(x+y)/(x^2+y^2)(x^2-y^2)÷1/(x^2+y^2)=(x+y)/(x^2-y^2)=(x+y)/(x+y)(x-y)=1/
=(x-4)/(x+1)(x-1)除以(x-4)(x+1)/(x+1)平方+1/(x-1)=(x-4)/(x+1)(x-1)乘以(x+1)平方/(x-4)(x+1)+1/(x-1)=(x-4)/(x+
x方-1分之x方+x化简2x(1+x)
x/(x²+2x+1),(x-1)/(x²+x),1/(x²-1)式1:x/(x+1)²,式2:(x-1)/[x(x+1)],式3:1/[(x+1)(x-1)]
(x²-1)/(x²+x)÷[x-(2x-1)/x]=(x+1)(x-1)/[x(x+1)]÷(x²-2x+1)/x=(x-1)/x*x/(x-1)²=1/(x
原式=[(x+2)/x(x-2)-(x-1)/(x-2)²]×x/(x-4)=[(x²-4-x²+x)/x(x-2)²]×x/(x-4)=[(x-4)/x(x-
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