方程3x 2(1−x)=4的解是( )
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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
∵x1,x2是方程2x2+3x-4=0的两个根,∴由韦达定理,得x1+x2=-32;x1•x2=-2;∴1x1+1x2=x1+x2x1•x2=−32−2=34,即1x1+1x2=34.
3/x2=1/x2-x即3*2-3x=x*22x*2=3Xx=0(舍去)x=3/2
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
等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了
(2x^2-4x-3)/(x^2-2x-1)-3=0{(2x^2-4x-3)-3(x^2-2x-1)}/(x^2-2x-1)-=0{-x^2+2x}/(x^2-2x-1)=0-x(x-2)/(x^2-
3x/(x+1)-(x+4)/(x^2+x)=-23x^2-(x+4)=-2(x^2+x)3x^2-x-4=-2x^2-2x5x^2+x-4=0(5x-4)(x+1)=0x1=4/5x2=-1经检验,
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
根据题意得x1+x2=-43,x1•x2=-53,所以1x1+1x2=x1+x2x1x2=−43−53=45,x12+x22=(x1+x2)2-2x1•x2=(-43)2-2×(-53)=469.故答
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-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
x²+2x+1=10(x+1)²=10x+1=3或x+1=-3所以x=2或x=-4【(x²+4)/x-4】÷【(x²-4)/(x²+2x)】=【(x&
方程x2+1x2−3x=3x-4可变形为:x2-3x+1x2−3x+4=0,∵y=x2-3x,∴y+1y+4=0,整理得:y2+4y+1=0.故选C.
题目写清楚点儿啊X1+X2=-3/2X1*X2=-2|X1-X2|=√41/2析:由根与系数的关系即得X1+X2=-3/2与X1*X2=-2而|X1-X2|^2=(X1+X2)^2-4X1*X2m=-
x2+x-12=0(x-3)(x+4)=0x=3或x=-4第二题:x2+x=4x+8x2-3x-8=0所以x1+x2=3x1x2=-8
∵一元二次方程3x2-4x-1=0的二次项系数a=3,一次项系数b=-4,常数项c=-1,∴x=−b±b2−4ac2a=4±16+122×3=2±73,∴x1=2+73,x2=2−73.
(5x+2-x+2)÷x-32x2+4x=(5x+2-x-21)•2x(x+2)x-3=5-(x+2)(x-2)x+2•2x(x+2)x-3=-(x+3)(x-3)x+2•2x(x+2)x-3=-2x
由题意只a不等于-2,如图