方程1 (x2-1=10 3x)实数根
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设x2-x=y,则1x2−x=1y,原方程化为y+1=6y,∴y2+y-6=0即(y+3)(y-2)=0,解得y1=-3,y2=2.当y=-3时,x2-x=-3,∴x2-x+3=0,∵△=1-12<0
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
3/x2=1/x2-x即3*2-3x=x*22x*2=3Xx=0(舍去)x=3/2
将原方程变形为x2+x+1x2+1+x2+1x2+x+1=23+32.设y=x2+x+1x2+1,则原方程变为y+1y=23+32,解得y1=23,y2=32.当x2+x+1x2+1=23时,x=-3
x^2-3x-1=0a=1b=-3 c=-1△=b^2-4ac=3^2-4*-1=13两个根为,所以X1=(3+根号13)/2 X2=(3-根号
设y=x2+x,则得y+1=2y,方程两边同乘以y,整理得y2+y-2=0.故本题答案为:y2+y-2=0.
令a=x2+x(a+1)(a+12)=42a2+13a+12=42a2+13a-30=0(a+15)(a-2)=0a=-15,a=2x2+x=-15x2+x+15=0无解x2+x=2x2+x-2=(x
7/(x+x2)-3/(x-x2)=6/(x2-1)两边同乘以x(x+1)(x-1),得7(x-1)+3(x+1)=6x7x-7+3x+3=6x10x-6x=3-74x=-4x=-1经检验x=-1是增
令x²+x=t原方程变为t+1=6/tt²+t-6=0(t+3)(t-2)=0则t=2或-31)x²+x=2x²+x-2=0(x+2)(x-1)=0x=-2或x
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
再答:这样再答:答案等于4/3颠倒一下我看错了再答:对不起啊
由原方程,得x2+(1x)2+5(x+1x)-66=0,∴x2+2+(1x)2+5(x+1x)-66-2=0,即(x+1x)2+5(x+1x)-68=0,∵设x+1x=t,∴原方程可化为t2+5t-6
(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-
把题拍过来帮你解
已知方程,然后x=±√7