先化简x方-2x 1分之x方-1
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(x+1)/x²-2x²/(x+1)=1设(x+1)/x²=t,则方程变为:t-2/t=1,即:t²-t-2=0,即(t-2)(t+1)=0∴t1=2t2=-1
x-x分之1=3x^2-2+1/x^2=9x^2+1/x^2=11x^4+1=11x^2x^4=11x^2-1x-x分之1=3x^2-1=3xx^2=3x+1x的四次方-2x的三次方-2x的二次方-7
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
自己想
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,
x方-5x+1=0两边除以x得∴x-5+1/x=0x+1/x=5两边平方得x²+2+1/x²=25x²+1/x²=23
x1+x2=5;x1x2=1;(1)x1/x2+x2/x1=(x1²+x2²)/(x1x2)=((x1+x2)²-2x1x2)/(x1x2)=(25-2)/1=23;(2
化简结果:-X的3次方分之一乘以(5X方-4分之1乘以X+2分之1)
(x^2+2x+1)/(x^2-3x-4)=(x+1)^2/(x+1)(x-4)=(x+1)/(x-4)
x^2-9=(x^2-6x+9)/22(x^2-9)=(x^2-6x+9)2(x-3)(x+3)-(x^2-6x+9)=02(x-3)(x+3)-(x-3)^2=0(x-3)[2(x+3)-(x-3)
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)………………
f(X)等于1+2X二次方分之X二次方f(a)=a²/(1+2a²)
=(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-1)/(x^2+3x+2)+6/(2+x-x^2)-(10-x)/(4-x^2)=(x-1)/(x+1)(x+2)-6/(x+1)(x-2)-(x-10)/(x+2)(x-2)=[(x-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)]
韦达定理x1+x2=mx1x2=2m-1所以x1²+x2²=(x1+x2)²-2x1x2=m²-4m+2=7m²-4m-5=0(m-5)(m+1)=0
1.原式=x-1分之x^2-(x+1)=(x-1分之x^2)-(x-1分之x^-1)=(x-1)分之1第二题可不可以写得再清楚一点,看起来好像有点乱诶~
-2或5又二分之一再问:过程?再答:2X²-3X-1=0(2X+1)(X-1)=0X1=-1/2,X2=1或者X1=1,X2=-1/2带到后面的式子就可以了