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用换元法解方程(x^2+1/x^2 )+5(x+1/x)%4+=0

化为: y+1/y-2=0 y²+1-2y=0 (y-1)²=0 y=1

2(x^2+1/x^2)-3(x+1/x)-5=0 设x+1/x=y 原式化为 2*(y^2-2)-3y-5=0 2y^2-3y-9=0 y=3或y=-3/2 y=3时 x+1/x=3 x^2-3x+1=0 x=(3±根号5)/2 y=-3/2时,x+1/x=-3/2 x^2+3/2x+1=0 无解 所以解为x=(3±根号5)/2

1、题目应是 2/(x-1)+4/(x^2-1)/4=0 2/(x-1)+4/(x-1)(x+1)=0 2/(x-1)=t x-1=2/t x+1=2/t +2 (x+1)/2=(1+t)/t 2/(x+1)=t/(1+t) 则方程变为 t+t*t/(1+t)=0 t(1+t)+t^2=0 2t^2+t=0 t(2t+1)=0 t=0(舍去) t=-1/2 2/(x-1)=-1/2 1-x=4 x=-3 经验证x...

这道题要求计算能力很强 3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0 [3/x+3/(x-5)]+[1/(x-1)+1/(x-4)]+[4/(x-2)+4/(x-3)]=0 (6x-15)/(x^2-5x)+(2x-5)/(x^2-5x+4)+(8x-20)/(x^2-5x+6)=0 (2x-5)[3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6)]=0 所...

y=x2x2?1,原方程可化为y-5×1y+1=0,去分母得,y2+y-5=0.故答案为:y2+y-5=0.

∵ x x-9 =y ,∴ ( x x-9 ) 2 -6( x x-9 )+5=0 可化为y 2 -6y+5=0,即(y-9)(y-5)=0,故选D.

∵x^-1/x-x/x^-1+2=0 y=x^-1/x ∴x^-1/x=y x/x^-1=1÷(x^-1/x)=1/y 则原式=y-1/y+2=0 y^2+2y-1 D 望采纳哦

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