计算:[(x/y-y/x)÷(x+y)+x(1/y-1/x)]÷y分之1+x

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计算:[(x/y-y/x)÷(x+y)+x(1/y-1/x)]÷y分之1+x
计算:[(x/y-y/x)÷(x+y)+x(1/y-1/x)]÷y分之1+x

计算:[(x/y-y/x)÷(x+y)+x(1/y-1/x)]÷y分之1+x
推荐答案的第一步运算就丢了大括号,改变了运算顺序和结果
当然另一个答案就偏离更远了,假设什么x/y=2,x=2y,这个条件题目并没有给出
正确解答为:
[(x/y-y/x)÷(x+y)+x(1/y-1/x)]÷[(1+x)/y]
=[(x²-y²)/xy(x+y) +x(x-y)/xy ]* y/(1+x)
=[(x-y)/xy+ x(x-y)/xy]*y/(1+x)
=[(x-y)(1+x)/xy]*y/(1+x)
=(x-y)(1+x)y/xy(1+x)
=(x-y)/x
=1-y/x

x/y=2
x=2y
[(x/y-y/x)÷(x+y)+(1/y-1/x)]÷(1+x)/y
=[(2-1/2)÷3y+1/y-1/2y]÷(1/y+2)
=(1/2y+1/y-1/2y)÷(1/y+2)
=1/y÷(1/y+2)
=1/(1+2y)
=1/(1+x)

[(x/y-y/x)÷(x+y)+x(1/y-1/x)]÷y分之1+x
=(x²-y²)/xy(x+y) +x(x-y)/xy * y/(1+x)
=(x-y)/xy+ (x-y)/(1+x)
=(x-y)[1/xy+1/(1+x)]
=(x-y)(x+1+xy)/[xy(1+x)]