将4,-3,4x-1,5x 2-搜答案-搜狗做题网

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

:(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&

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/

∵A=5x2+4x-1，B=-x2-3x+3，C=8-7x-6x2，∴A-B+C=（5x2+4x-1）-（-x2-3x+3）+（8-7x-6x2）=5x2+4x-1+x2+3x-3+8-7x-6x2=

等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了

（1）4x2-3x=52移项得，4x2-3x-52=0，（x-4）（4x+13）=0，x1=4，x2=-134．（2）5x2=4-2x原方程可化为5x2+2x-4=0，a=5，b=2，c=-4，△=4

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

=1/(x-2)(x-1)+1/(x-2)(x-3)+1/(x-3)(x-4)-4/(x-1)(x-5)帮你删除了1/x2-7=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x

（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

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

原式=[x+2x(x-2)-x-1(x-2)2]÷x2-16x2+4x=[x2-4x(x-2)2-x2-xx(x-2)2]÷x2-16x2+4x=x-4x(x-2)2•x(x+4)(x+4)(x-4)

原式=x3+5x2+4x-1+x2+3x-2x3+3+8-7x-6x2+x3=10，故与x无关．

[(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=2x2-1，当x=-3时，原式=2×（-3）2-1=17．

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：