1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy4.(1-1-x分之1)÷1-x分之1+x分式的加减法
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/30 12:55:05
![1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy4.(1-1-x分之1)÷1-x分之1+x分式的加减法](/uploads/image/z/3000904-16-4.jpg?t=1.a%2B1%E5%88%86%E4%B9%8B1-1-a%E5%88%86%E4%B9%8B2+2.x-y%E5%88%86%E4%B9%8Bx-x%2By%E5%88%86%E4%B9%8By+3.%EF%BC%88x%2By%E5%88%86%E4%B9%8Bx%26%23178%3B-x%2By%E5%88%86%E4%B9%8By%26%23178%3B%EF%BC%89%C2%B7x-y%E5%88%86%E4%B9%8Bxy4.%EF%BC%881-1-x%E5%88%86%E4%B9%8B1%29%C3%B71-x%E5%88%86%E4%B9%8B1%2Bx%E5%88%86%E5%BC%8F%E7%9A%84%E5%8A%A0%E5%87%8F%E6%B3%95)
1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy4.(1-1-x分之1)÷1-x分之1+x分式的加减法
1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy
4.(1-1-x分之1)÷1-x分之1+x
分式的加减法
1.a+1分之1-1-a分之2 2.x-y分之x-x+y分之y 3.(x+y分之x²-x+y分之y²)·x-y分之xy4.(1-1-x分之1)÷1-x分之1+x分式的加减法
解1题:
原式=[1/(a+1)]-[2/(1-a)]
=[1/(a+1)]+[2/(a-1)]
={ (a-1)/[(a+1)(a-1)] }+{2(a+1)/[(a+1)(a-1)] }
=[(a-1)+2(a+1)]/[(a+1)(a-1)]
=(a-1+2a+2)/(a²-1)
=(3a+1)/(a²-1)
解2题:
原式=[x/(x-y)]-[y/(x+y)]
={ x(x+y)/[(x+y)(x-y)] }-{ y(x-y)/[(x+y)(x-y)] }
=[x(x+y)-y(x-y)]/[(x+y)(x-y)]
=(x²+xy-xy+y²)/(x²-y²)
=(x²+y²)/(x²-y²)
解3题:
原式=[x²/(x+y)-y²/(x+y)]×[xy/(x-y)]
=[(x²-y²)/(x+y)]×[xy/(x-y)]
=[(x+y)(x-y)/(x+y)]×[xy/(x-y)]
=xy
解4题:
原式=[1-1/(1-x)]÷[(1+x)/(1-x)]
={ [(1-x)-1]/(1-x) }×[(1-x)/(1+x)]
=[-x/(1-x)]×[(1-x)/(1+x)]
=-x/(1+x)