6.8x 2.8x=1.92解方程

X1,X2为方程x²+3x+1=0的两根那么x1²+3x1+1=0x1²=-3x1-1x1（-3x1-1）+8x2+20=-3x1²-x1+8x2+20=-3（

X1.X2是方程:X的平方+3X+1=0的两个实数根则：X1²+3X1+1=0X1²=-3X1-1由韦达定理得：X1+X2=-3X1的三次方+8*X2+20=X1*X1²

x1三次方+8X2+20=x1³+3x1²+x1+8x2+20-3x1²-x1=x1(x1²+3x1+1)+8x2+20-3x1²-x1………………x

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&#

x1带入方程得：x1²+3x1+1=0再同乘上x1得：x1³+3x1²+x1=0所以x1³=-3x1²-x1=-3(-3x1-1)-x1=8x1+3所

x1=(-3+√5）/2；x2=(-3-√5)/2若X1=x1,X2=x2,则X1²+8X2+20=(14-6√5）/4-4(3+√5)+20=7/2-12+20-3√5/2-4√5=23/

x=7(x²-x+1)7x²-7x+7=x7x²+7=8x两边平方49x4+98x²+49=64x²两边减去49x²49x4+49x&sup

X1、X2是方程X^2+3X+1=0的两实数根韦达定理得:X1+X2=-3X1X2=1X1^2+3X1+1=0x1^2=-(3x1+1)x1^3+8x2+20=-x1*(3x1+1)+8x2+20=-

原式=1/(1-x)+1/(1+x)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)-1/(1-x)添一项减一项=2/(1-x^2)+2/(1+x^2)+4/(1+x^4)+8/(1+x^

x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x

1、y=2x²-6x由题,y=2（x²-3x）=2[x²-3x+（3/2）^2-（3/2）^2]=2(x-3/2)^2-2*9/4=2(x-3/2)^2-9/2所以：y=

X/(X^2+X+1)=a1/(x+1+1/x)=ax+1/x=1/a-1x^2+1/x^2+2=(1-a)^2/a^2x^2+1/x^2+1=(1-a)^2/a^2-1=(1-2a)/a^2X^2/

可以由十字相乘法分解因式为（3x-8)(x+1)=0,解得x1为-1,x2为8/3再问：完整可以吗

x+1/x=4两边乘xx²+1=4x两边平方x^4+2x²+1=16x²两边减去x²x^4+x²+1=15x²所以x²/（x^+x

由韦达定理得x1+x2=3,x1*x2=3/2则x1²+x2²=(x1+x2)²-2x1*x2=9-3=6x1³+x2³=(x1+x2)(x1

由x2(x的二次方)+x-1=0得出x2(x的二次方)+x=1那么：x3(x的三次方)+2x2(2x的二次方)-3=x3(x的三次方)+x2(x的二次方)+x2(x的二次方)-3=x【x2(x的二次方

-1x1+x2=-3x1^2=-3x1-1x1^3+8x2+20=x1(-3x1-1)+8x2+20=-3x1^2-x1+8x2+20=-3(-3x1-1)-x1+8x2+20=9x1+3-x1+8x

2x^5-5x^4+2x^3-8x^2+3x=2x^5-6x^4+2x^3+x^4-3x^3+x^2+3x^3-9x^2+3x=2x^3(x^2-3x+1)+x^2(x^2-3x+1)+3x(x^2-

x²+1/x²=3两边平方x的四次方+2+1/x的四次方=9x的四次方+1/x的四次方=7

-2或5又二分之一再问：过程？再答：2X²-3X-1=0(2X+1)(X-1)=0X1=-1/2,X2=1或者X1=1,X2=-1/2带到后面的式子就可以了