字典翻译 问答 小学 数学 【分式的加减】计算题(1)(x+y/x-y)+(2x/y-x)=(2)(2x/x²-y²)+(3x/y²-x²)+【x/(x+y)(y-x)】=(3)(x²-x-2/x+3)÷(2-x)+(x+1/x+3)=
问题标题:
【分式的加减】计算题(1)(x+y/x-y)+(2x/y-x)=(2)(2x/x²-y²)+(3x/y²-x²)+【x/(x+y)(y-x)】=(3)(x²-x-2/x+3)÷(2-x)+(x+1/x+3)=
问题描述:

【分式的加减】计算题

(1)(x+y/x-y)+(2x/y-x)=

(2)(2x/x²-y²)+(3x/y²-x²)+【x/(x+y)(y-x)】=

(3)(x²-x-2/x+3)÷(2-x)+(x+1/x+3)=

林成武回答:
  1)(x+y/x-y)+(2x/y-x)   =(x+y)/(x-y)-2x/(x-y)   =(x+y-2x)/(x-y)   =(y-x)/(x-y)   =-1   (2)(2x/x²-y²)+(3x/y²-x²)+【x/(x+y)(y-x)】   =2x/(x²-y²)-3x/(x²-y²)-x/(x²-y²)   =(2x-3x-x)/(x²-y²)   =-2x/(x²-y²)   (3)(x²-x-2/x+3)÷(2-x)+(x+1/x+3)   =(x²-x-2)/(x+3)(2-x)+(x+1)/(x+3)   =(x²-x-2)/(x+3)(2-x)+(x+1)(2-x)/(x+3)(2-x)   =(x²-x-2+2x+2-x²-x)/(x+3)(2-x)   =0/(x+3)(2-x)   =0
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