作业帮 > 数学 > 作业

计算:1/(m²-m)+(m-2)/(2m²-2),并求当m=3时,原式的值

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/30 18:59:01
计算:1/(m²-m)+(m-2)/(2m²-2),并求当m=3时,原式的值
还有一题,
用两种不同的运算顺序计算
[x/(x-2)-x/(x+2)]×[(2-x)/x]
急啊,记住,先化简再求值
1/(m²-m)+(m-2)/(2m²-2)
=1/m(m-1)+(m-2)/2(m+1)(m-1)
=[2(m+1)+(m-2)m]/2m(m+1)(m-1)
=(2m+2+m²-2m)/2m(m+1)(m-1)
=(m²+2)/2m(m+1)(m-1)
m=3代入得
(9+2)/(2*3*4*2)
=11/48
先通分再算乘
[x/(x-2)-x/(x+2)]×[(2-x)/x]
=[x(x+2)-x(x-2)]/(x-2)(x+2)×[-(x-2)/x]
=-(x²+2x-x²+2x)/[(x+2)/x]
=-4x/[(x+2)/x]
=-4/(x+2)
直接用分配率
[x/(x-2)-x/(x+2)]×[(2-x)/x]
=x/(x-2)*[(2-x)/x]-x/(x+2)* (2-x)/x
=-1-(2-x)/(x+2)
=(-x-2+x-2)/(x+2)
=-4/(x+2)