x的平方 4x 2=0,那么3x的平方 12x 2010-搜答案-智慧作业帮

由y=x2+3x，则（x2+3x）2+2（x2+3x）-3=0，可化为：y2+2y-3=0，分解因式，得，（y+3）（y-1）=0，解得，y1=-3，y2=1，故选B．

∵x1,x2是方程2x^2+3x-4=0的两个根∴x1+x2=-b/a=-3/2,x1x2=c/a=-4/2=-2∴|x1-x2|=√(x1-x2)^2=√[(x1+x2)^2-4x1x2]=√[(-

X1,X2是方程2x的平方+3x-4=0的两个实数根x1+x2=-3/2x1x2=-2x1^2+2x1x2+x^2=9/4x1^2-2x1x2+x^2=9/4-4x1x2(x1-x2)^2=41/4x

X2平方+2X-3=0X2平方+2X=33X2的平方+6X-4=3（X2平方+2X）-4=3*3-4=5

(X1)³-4(X2)²+19=(X1)*(X1)²-4(X2)²+19=(X1)*(-X1+3)-4(-X2+3)+19=-(X1)²+3(X1)+

x^2-3x+2=0(x-2)(x-1)=0x=2或x=1当x=2时x^2+1/x^2=2^2+1/2^2=4+1/4=17/4当x=1时x^2+1/x^2=1^2+1/1^2=1+1=2

用韦达定理.x1-x2=根号（(x1+x2)^2-4x1x2),韦达定理代入相应数值即可

根据韦达定理x1+x2=-3/2,x1x2=-2所以x1²+x2²=(x1+x2)²-2x1x2=(-3/2)²+4=9/4+4=25/4

x的3次-2x+10=x³+x²-x-x²-x+1+9=x(x²+x-1)-(x²+x-1)+9∵x²+x-1=0∴原式=9

设x1>x2,x1+x2=-1对方程变形：x1^2+x1-3=0x1^3+x1^2-3x1=0x2^2+x2-3=0所以x1^3-4x2^2+19=3x1-x1^2-4(3-x2)+19=3x1-(3

原式=-12x2+5x+8用解二次方程的式子代入,得X1=0.5X2=-1/12

x1+x2=3/2x1x2=m/21.△=9-8m>=0,∴m0,∴m>0∴0

x1+x2=-2,x1*x2=-32x1（x2^2+5x2-3）=2x1*x2^2+10x1*x2-6x1=-6x2-30-6x1=-6（x1+x2）-30=12-30=-18再问：第二步中x2的平方

x2-3x+1=0两边同时除以X,得X-3+1/X=0移项,得X+1/X=3接着两边同时平方,得（X+1/X)^2=3^2展开X^2+2*X*1/X+1/X^2=9X^2+2+1/X^2=9X^2+1

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+4x-2=0得x²+4x=2而3x²+12x+2004=3（x²+4x）+2004=3×2+2004=2010

X1+X2＝-3/2X1X2=-1/21/X1+1/X2=(X1+X2)/X1X2=3

4x前面是+还是-?再问：+再答：∵x²+4x-2=0∴x²+4x=2∴3x²+12x=6∴3x²+12x+2012=6+2012=2018

X²－4X＋3≤0(X-1)(X-3)≤01≤X≤3∵x∈zX为正整数∴X=｛1,2,3｝

x1+x2=-3/2x1x2=-21/x1+1/x2=（x1+x2)/x1x2=(-3/2)/(-2)=3/4x1²+x2²=(x1+x2)²-2x1x2=(-3/2)&