x-4＝29 1 2x 2＝6 3x 1＝4 4x-2＝2-搜答案-百度作业网

∵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=-1x1x2=-4原式=x1(4-x1)-5(4-x2)+10=4x1+x1-4+5x2-10=5(x1+x2)-14=-5-14=-19

x1+x2=4/5x1x2=-1/5所以1/x1+1/x2=(x1+x2)/(x1x2)=-4(x1+x2)²=(4/5)²x1²+2x1x2+x2²=16/2

如图所示,条件区间为途中阴影部分.Z=x1+3x2的斜率=-1/3,Z为函数与Y轴交点的纵坐标.由图可知,当函数过点A时Z最大,求的A坐标为（2,4）,代入Z=x1+3x2得Z=14所以最大值为14有

x^2-4x+3=0(x-3)(x-1)=0x1=3,x2=1x1*x2=3

根据韦达定理有X1+X2=-b/a=-2/3,X1*X2=c/a=-3/3=-1①x2/x1+x1/x2=(x2²+x1²)/（x1x2）=【（x1+x2）²-2x1x2

f'(x)=3x²-4令f'(x)≥03x²-4≥03x²≥4x≥2/√3或x≤-2/√3即函数在区间(-∞,-2/√3]上单调递增；在区间[-2/√3,2/√3]上单调

x1+x2=4x1x2=k-3x1=3x2所以x1+x2=3x2+x2=4所以x2=1x1=3k=x1x2+3=6

解析原式：（x1x2)²-(x1+x2)=115根据韦达定理x1x2=c/a=kx1+x2=-b/a=6所以：（x1x2)²-(x1+x2)=115k²-6=115k=1

有韦达定理得x1+x2=2mx1*x2=m+2则(x1)²x2+x1(x2)²=x1*x2(x1+x2)=2m(m+2)=0解得m=0或-2当m=-2时,x^2+4x=0,有两个实

x1,x2是方程的根,所以满足x1²-x1-4=0,x2²-x2-4=0x1³-x1²-4x1=0,所以x1³=x1²+4x1x1³

证明：判别式=p^2-4p则x=-p±√p²-4q/2不妨设x1=-p+√p²-4q/2,x2=-p-√p²-4q/2∴x1+x2=-p+√p²-4q-p-√p

根据韦达定理得x1+x2=(4k-7)/9x1x2=-2k²/3若k=0则是9x²+7x=0x1=0,x2=-7/9,不符合|x1/x2|=3/2所以x1x2

将式子通分得（x1²+x2²）÷x1x2=1.5,再整理得[（x1+x2）²－2x1x2]÷x1x2=1.5,而根据维达定理知x1+x2=2－k、x1x2=k－2,求出k

X1X2＝1X1+X2＝3x1^2-4x1-x2=x1^2-4x1-(3-x1)=x1^2-3x1-3∵x1,x2是方程x^2-3x+1=0的解∴x1^2-3x1+1-4=-4

∵x1、x2为方程x＾2－2x－4＝0的两个根∴x1²-2x1-4=0;x1²=2x1+4x1+x2=2（x1）＾2－3（x1）－（x2）=2x1+4-3x1-x2=4-(x1+x

AB是抛物线Y方＝4X的焦点弦设为：y=k(x-1)则：y1+y2=k(x1+x2)-2k=6k-2k=4ky1^2=4x1,y2^2=4x2(y1^2-y2^2)=4(x1-x2)(y1-y2)/(

列增广矩阵,化为阶梯阵,选定基础解系,解出基础解系和特解.