作业帮1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)其中X的平方+3X+1=0
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/06 08:23:27
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)其中X的平方+3X+1=0
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)其中X的平方+3X+1=0
1/X(X+1)+1/(X+1)(X+2)+1/(X+2)(X+3)其中X的平方+3X+1=0
第一题
由X^2+3X+1=0,得X^2=-3X-1代入,
1/(X+1)(X+2)=1,故原式=1/X(X+1)+1+1/(X+2)(X+3)=
1+1/(-1-2X）+1/(2X+5)通分,原式=1+4/-(2X+5)(2X+1)=
1+4/-[4(X^2+3X+1)+1]=1-4=-3
第二题
移项,1/(x-10)-1/(x-7)=1/(x-9)-1/(x-6),通分得
3/(x^2-17x+70)=3/(x^2-15x+54),即(x^2-17x+70)=(x^2-15x+54),
得x=8
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分解因式----（X*X-X)(X*X-X-2)+1
】x(x+1)+x^2(x-1)
1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x*x)+8/(1-x*x*x*x*x*x*x*x)怎么做
x/x+(1-x)/x-1/x
x-1/x-2
{2X+1}+{X}
(x^2-1)/x
x^2-1/x
2X+|X+1|
｜x-1｜+｜x-2｜
(x+1)/(x+2)
/X+1/+/X+2/
2x/(1-x)
|X+2|+|X-1|
2x-1 (x