分式x x分之一=3
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1/x-1/y=3等号左边通分,得(y-x)/(xy)=3故x-y=-3xy代入分式[2x+3xy-2y]/[x-2xy-y]=[2(x-y)+3xy]/[(x-y)-2xy]=(-6xy+3xy)/
a+1/a=3a^2+1=3a平方(a^2+1)^2=9a^2a^4+2a^2+1=9a^2a^4+a^2+1=8a^2a^2/(a^4+a^2+1)=1/8
1/X+1/Y=2两边同乘以XY,得Y+X=2XY,于是,XY=(X+Y)/2(2X+3XY+2Y)/(X-XY+Y)=7楼上的刚好写反了.
6(x+1)的平方
方程两边都乘(x-3),得x-2(x-3)=m2∵原方程有增根,∴最简公分母x-3=0,即增根是x=3,把x=3代入整式方程,得m=±3.
(1)原式两边同时乘(x+2)(x-2),得2x(x-2)-3(x+2)=2(x+2)(x-2),2x2-4x-3x-6=2x2-8,-7x=-2,x=27.经检验x=27是原方程的根.(2)原式两边
原式=5x(x+3)(x−2)+2x−5(x+3)(x−4)-7x−10(x−2)(x−4)=5x(x−4)+(2x−5)(x−2)−(7x−10)(x+3)(x+3)(x−2)(x−4)=−40x+
1/x--1/y=3,化简y--x=3xy(2x-3xy-2y)/(x-2xy-y)=[-2(y-x)-3xy]/(x-y-2xy)=-9xy/-5xy=9/5
xy=x-y两边同时除以xy1=1/y-1/xy分之一减x分之一等于1
是,但不是分式方程分式是指分母中有未知数的式子
[2x+14xy-2y]/[3x-2xy-3y]=(2/y+14-2/x)/(3/y-2-3/x)分子分母同时除以xy=[-2(1/x-1/y)+14]/[-3(1/x-1/y)-2]=(-6+14)
x+x分之一=3两边平方:x²+1/x²+2=9x²+1/x²=9-2=7
方程两边都乘以(x-1)(x+1),得x(x+1)+k(x+1)-x(x-1)=0.解得x=-kk+2,∵分式方程无解,∴−kk+2=±1,解得k=-1,故答案为:-1.
1/x-1/y=3(y-x))/xy=3所以x-y=-3xy原式=[2(x-y)+3xu]/[(x-y)-2xy]=[2(-3xy)+3xu]/[(-3xy)-2xy]=-3xy/(-5xy)=3/5
1/x+5+1/x+8=1/x+6+1/x+71/x+5-1/x+6=1/x+7-1/x+8(x+6)/[(x+5)(x+6)]-(x+5)/[(x+5)(x+6)]=(x+8)/[(x+7)(x+8
x+1/x=3两边平方x^2+2*x*1/x+1/x^2=3^2x^2+2+1/x^2=9x^2+1/x^2=7x^2/(x^4+x^2+1)上下除以x^2=1/(x^2+1+1/x^2)=1/(7+
分式4−xx−3无意义,则x-3=0,解得x=3;分式|x|−9x+9的值等于零,则|x|-9=0且x+9≠0,解得x=9.故答案为3,9.
分母有未知数所以是分式再问:两个分式相加还是分式?再答:是的只要分母有未知数就是分数,不管有没有加减乘除的符号
方程两边都乘(x-3),得x-2(x-3)=k,∵原方程有增根,∴最简公分母x-3=0,即增根是x=3,把x=3代入整式方程,得k=3.
1/x-1/y=(y-x)/xy=-(x-y)/(x-y)=-1