3x-6=x 2怎么求方程

利用两根和、两根积公式得x1+x2=-2/3,x1x2=-6/3=-2x1*x1+x1x2+x2*x2=x1*x1+2x1x2+x2*x2-x1x2=(x1+x2)^2-x1x2=(-2/3)^2+2

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

韦达定理x1+x2=-3/2,x1x2=-1/2

原式可化为1/(x+1)(x-3)+2/(x-3)((x+2)+3/(x+1)(x+2)=0两边同乘以：(x+1)(x+2)(x-3)得：(x+2)+2(x+1)+3(x-3)=0(x≠-1,x≠-2

∵x²+6x+3=0∴x1+x2=-6x1x2=3x1/x2+x2/x1=(x1+x2)²-2x1x2/x1x2=10

等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了

7/(X2+X)+3/(X2-X)=6/(X2-X),去分母,等式两端同时乘X(X+1)(X-1):7(X-1)+3(X+1)=6(X+1),7X-7+3X+3=6X+6,7X+3X-6X=6+7-3

方程4x^2-7x-3=0的两根为x1,x2,所以x1+x2=7/4,x1x2=-3/4,x2/(x1+1)+x1/(x2+1)=（x1^2+x2^2+x1+x2）/(x1x2+x1+x2+1)x1^

方程3x²-4x=-1可化为：3x²-4x+1=0由根与系数的关系,有x1+x2=4/3,x1x2=1/3∴x2/x1+x1/x2=(x1²+x2²)/(x1x

（x2-x）=3(x2+x)x2-x=3x2+3xx2-3x2-x-3x=0-2x2-4x=0-2x(x+2)=0x1=0x2=-2

(2x-1)(x-3)=0x1=1/2x2=3

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

X的平方吧!x1分之x2加x2分之1=x1x2分之x1的平方+x2的平方=x1x2分之（x1+x2)的平方-2x1x2=因为x1+x2=-6x1x2=3所以原式等于3分之30=10

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/(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

∵x²-3x+5+6/(x²-3x)=0∴设x²-3x=t则原方程变换为t+5+6/t=0==>t²+5t+6=0==>(t+2)(t+3)=0∴t=-2,或t

x1+x2=-3x1x2=-1所以x2／x1+x1／x2=(x2^2+x1^2)／x1x2=[(x1+x2)^2-2x1x2]/x1x2=（9+2）/（-1）=-11x2／x1*x1／x2=1所以方程

∵x1、x2是方程x2+6x+3=0的两实数根，∴由韦达定理，知x1+x2=-6，x1•x2=3，∴x2x1+x1x2=(x1+x2)2−2x1•x2x1•x2=(−6)2−2×33=10，即x2x1

已知X1X2为方程5X平方-3X-1=0两个根；所以x1+x2=3/5;x1x2=-1/5;x1-x2=√(x1-x2)²=√[(x1+x2)²-4x1x2]=√(9/25+4/5