Ответ:
a)[tex](\frac{b}{a+b} -\frac{b-a}{b}) :\frac{a}{b}=\frac{b}{a+b} *\frac{b}{a} -\frac{b-a}{b} *\frac{b}{a} = \frac{b^{2} }{(a+b)a} -\frac{b-a}{a}=\frac{b^{2}-b^{2}+a^{2} }{(a+b)a}=\frac{a}{a+b}[/tex]
б)[tex]\frac{3a}{1+c} -\frac{4}{1-c^{2} } *\frac{a-ac}{2} =\frac{3a}{1+c} -\frac{4a(1-c)}{(1-c)(1+c)2} =\frac{3a}{1+c}-\frac{2a}{1+c} =\frac{a}{1+c}[/tex]
в)[tex]\frac{x}{x-y} -\frac{y}{x+y} +\frac{2xy}{x^{2}-y^{2} } =\frac{x(x+y)-y(x-y)+2xy}{x^{2}- y^{2} } =\frac{(x+y)^{2} }{(x+y)(x-y)} =\frac{x+y}{x-y}[/tex]
г)[tex]\frac{m^{2}-n^{2} }{mn} :\frac{m-n}{3n} *\frac{1}{m+n} =\frac{3}{m}[/tex]
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Answers & Comments
Ответ:
a)[tex](\frac{b}{a+b} -\frac{b-a}{b}) :\frac{a}{b}=\frac{b}{a+b} *\frac{b}{a} -\frac{b-a}{b} *\frac{b}{a} = \frac{b^{2} }{(a+b)a} -\frac{b-a}{a}=\frac{b^{2}-b^{2}+a^{2} }{(a+b)a}=\frac{a}{a+b}[/tex]
б)[tex]\frac{3a}{1+c} -\frac{4}{1-c^{2} } *\frac{a-ac}{2} =\frac{3a}{1+c} -\frac{4a(1-c)}{(1-c)(1+c)2} =\frac{3a}{1+c}-\frac{2a}{1+c} =\frac{a}{1+c}[/tex]
в)[tex]\frac{x}{x-y} -\frac{y}{x+y} +\frac{2xy}{x^{2}-y^{2} } =\frac{x(x+y)-y(x-y)+2xy}{x^{2}- y^{2} } =\frac{(x+y)^{2} }{(x+y)(x-y)} =\frac{x+y}{x-y}[/tex]
г)[tex]\frac{m^{2}-n^{2} }{mn} :\frac{m-n}{3n} *\frac{1}{m+n} =\frac{3}{m}[/tex]