x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4如何分解因式
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 16:38:48
![x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4如何分解因式](/uploads/image/z/6918661-37-1.jpg?t=x%2B2%2Fx%2B1-x%2B3%2Fx%2B2-x-4%2Fx-3%2Bx-5%2Fx-4%E5%A6%82%E4%BD%95%E5%88%86%E8%A7%A3%E5%9B%A0%E5%BC%8F)
x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4如何分解因式
x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4如何分解因式
x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4如何分解因式
原式=(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=[(x+2)*(x+2)-(x+3)*(x+1)]/(x+1)*(x+2)--[(x-4)*(x-4)-(x-3)*(x-5)]/(x-3)*(x-4)
=1/(x+1)*(x+2)-1/(x-3)*(x-4)
=[(x-3)*(x-4)-(x+1)*(x+2)]/[(x+1)*(x+2)*(x-3)*(x-4)]
=10(1-x)/[(x+1)*(x+2)*(x-3)*(x-4)]
-x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4
=-4/X-4=-4(1+1/X)
扬帆知道快乐10(x^2+2x+3)/[(x+1)(x+2)(x-3)(x-4)].
x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4
解:
原式=[1+ 1/(x+1)] - [1+ 1/(x+2)] - [1- 1/(x-3)] +[1- 1/(x-4)]
= 1/(x+1) -1/(x+2) +1/(x-3) -1/(x-4)
=[(x+2)-(x+1)] /[(x+1)(x+2)] +[(x-4)-(x-3)] /[(x-3)(...
全部展开
x+2/x+1-x+3/x+2-x-4/x-3+x-5/x-4
解:
原式=[1+ 1/(x+1)] - [1+ 1/(x+2)] - [1- 1/(x-3)] +[1- 1/(x-4)]
= 1/(x+1) -1/(x+2) +1/(x-3) -1/(x-4)
=[(x+2)-(x+1)] /[(x+1)(x+2)] +[(x-4)-(x-3)] /[(x-3)(x-4)]
=1/[(x+1)(x+2)] - 1/[(x-3)(x-4)]
=([(x-3)(x-4)) -[(x+1)(x+2)]) /[(x+1)(x+2)(x-3)(x-4)]
=[(x^2 -7x+12) -[x^2+3x+2)]/[(x+1)(x+2)(x-3)(x-4)]
= (-10x+10)/[(x+1)(x+2)(x-3)(x-4)]
=10(1-x)/[(x+1)(x+2)(x-3)(x-4)]
我刚做过过这道题。。。。
收起