将4,-3,4x-1,5x 2两两-搜答案-拍题作业网

2x1²+4x2²-6x2+2011=2x1²+2x2²+2x2²-6x2+2011=2(x1²+x2²)+2(x2²-

(x²+1)²-4x(x²-1)=（x²-1）²-4x(x²-1)+4x²=（x²-1-2x）²（x^4-2x

圆系方程x^2+y^2-x-y-2+a(x^2+y^2+4x-4y)=0代入点(3,1)9+1-3-1-2+a(9+1+12-4)=04+a(18)=0a=-2/9x^2+y^2-x-y-2-2/9*

:(1)2x2-5x+x2+4x,其中x=-3=3x²-x=3x(-3)²+3=27+3=30(2)(3x2-xy-2y2)-2(x2+xy-2y2),其中x=6,y=-1=3x&

x²－4x＋2＝0由韦达定理得：x1＋x2＝4,x1·x2＝2∴(1)x1+x2+3x1x2=4+3*2=10(2)x2/x1+x1/x2=(x2²+x1²)/x1x2=

1/(x²+3x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1)-1/(x+2)同理1/(x²+5x+6)=1/(x+2)-1/(x+3)1/(x²

1/(x2-5x+6)-1/(4x-x2-3)-1/(3x-x2-2)=1/(x2-5x+6)+1/(x2-4x+3)+1/(x2-3x+2)=1/(x-2)(x-3)+1/(x-3)(x-1)+1/

等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了

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

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^

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

（x2-x-6）（x2+3x-4）+24=（x-3）（x+2）（x-1）（x+4）+24=（x-3）（x+4）（x-1）（x+2）+24=（x2+x-12）（x2+x-2）+24=（x2+x）2-14

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

1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=1/x-1/

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

原式=（5-3-2）x2+（-5+6）x+（4-5）=x-1

[(x-4)/(x^2-1)]÷[(x^2-3x-4)/(x^2+2x+1)]+1/(x-1)=[(x-4)/(x-1)(x+1)]÷[(x+1)(x-4)/(x+1)^2]+1/(x-1)=[(x-

原式=5x²-x²-(4x-x²)+2(x²-3x)=4x²-4x+x²+2x²-6x=7x²-10x

3（x+1）^2+10(x+1)+14

解题思路：这个是因式分解问题。由完全平方公式，再应用换元法可以得到结果.解题过程：