解方程1-2 x-1 x2=2x x2
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设(x²-1)/(x²+2x)=t则8t+3/t=118t²-11t+3=0(8t-3)(t-1)=0解得t=3/8或t=11.t=3/8(x²-1)/(x
后面的x²+11x-708有误吧!再问:没有题目就这样能不能帮我再答:那我就试试:原式为:1/x2+x+1/x2+3x+2+1/x2+5x+6+1/x2+7x+12+1/x2+9x+20=5
2/(x2-x)+6/(1-x2)=7/(x2+x)2/x(x-1)-6/(x-1)(x+1)=7/x(x+1)[x*(x-1)*(x+1)]*[2/x(x-1)-6/(x-1)(x+1)]=[7/x
(1)去分母得:x(x+2)+2(x-2)=x2-4,去括号得:x2+2x+2x-4=x2-4,移项合并得:4x=0,解得:x=0,经检验x=0是分式方程的解;(2)原式=[x+2x(x−2)-x−1
方程两边都乘x(x+1),得2x2-(m+1)=(x+1)2∵原方程有增根,∴最简公分母x(x+1)=0,解得x=0或-1,当x=0时,m=-2.当x=-1时,m=1,故选D.
(1)方程两边同乘(x-1)(x+1),得:2(x-1)-x=0,整理解得x=2.经检验x=2是原方程的解.(2)方程两边同乘(x-3)(x+3),得:3(x+3)=12,整理解得x=1.经检验x=1
∵(x+1x)2=x2+1x2+2,∴方程:2(x2+1x2)-3(x+1x)-1=0可化为2(x+1x)2−3(x+1x)−5=0.因式分解为[2(x+1x)−5][(x+1x)+1]=0,∴2(x
方程左边1/(xx+x)+1/(xx+3x+2)+1/(xx+5x+6)+1/(xx+7x+12)+1/(xx+9x+20)对分母因式分解得1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(
去分母:(2-x)=x-3+1,化简得:2x=4,∴x=2,经检验,原分式方程的根是:x=2.
x2+x+1=2/(x2+x)(X²+x)²+(x²+x)-2=0(x²+x+2)(x²+x-1)=0∴x²+x-1=0x=(-1±√5)/
两边乘x(x+1)(x-1)2(x-1)+3(x+1)=4x2x-2+3x+3=4x5x+1=4xx=-1经检验,x=-1时分母x+1=0增根,舍去方程无解
(x²+x)(x²+x-2)=-1把(x²+x)看成整体(x²+x)[(x²+x)-2]=-1运用乘法分配率(x²+x)²-2(x
x²+x-1/(x²+x)=3/2两边同时乘以(x²+x)得:(x²+x)²-1=3(x²+x)/22(x²+x)²-3
通分:(x^2-3x)+(2x-1)(x+1)=0化简:3x^2-2x-1=0x1=1(舍去,分母不为0)x2=-1/3
(x-4)/(x²+x-2)=1/(x-1)+(x-6)/(x²-4)(x-4)/(x-1)(x+2)=1/(x-1)+(x-6)/(x-2)(x+2)(x-4)(x-2)=(x-
x²-2x=2x+1x²-4x=1x²-4x+4=5(x-2)²=5x-2=±√5x=2±√5
2/x2=30=0x=±1/√15再问:过程再答:合并同类项1/x2+1/x2+1/x2+2x+11x-13x+10+10+10=03/x2+30=010x2+1=0x==±√10i/10刚才答案有误
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已知方程,然后x=±√7