Задание 535

Решение:

• \({\largeа)}\ \left(\frac{a+x^{\backslash{x}}}{a\phantom{^{\backslash{x}}}}-\frac{x-y^{\backslash{a}}}{x\phantom{^{\backslash{a}}}}\right)\cdot\frac{a^2}{x^2+ay}=\frac{ (a+x)\cdot{x}-(x-y)\cdot{a}}{ax}\cdot\frac{a^2}{x^2+ay}=\frac{ax+x^2-ax+ay}{ax}\cdot\frac{a^2}{x^2+ay}=\frac{x^2+ay}{ax}\cdot\frac{a^2}{x^2+ay}=\frac{1}{x}\cdot\frac{a}{1}=\frac{a}{x}\)

• \({\largeб)}\ \left(\frac{a}{a-1}+1^{\backslash{a\ -\ 1}}\right):\left(1^{\backslash{a\ -\ 1}}-\frac{a}{a-1}\right)=\frac{ a+1\cdot(a-1)}{a-1}:\frac{ 1\cdot(a-1)-a}{a-1}=\frac{a+a-1}{a-1}:\frac{a-1-a}{a-1}=\frac{2a-1}{a-1}:\frac{{-}1}{a-1}=\frac{2a-1}{a-1}\cdot\left({-}\frac{a-1}{1}\right)={-}(2a-1)={-}2a+1\)

• \({\largeв)}\ \left(m^{\backslash{1\ +\ m}}-\frac{1}{1+m}\right)\cdot\frac{m+1}{1-m-m^2}=\frac{ m\cdot(1+m)-1}{1+m}\cdot\frac{m+1}{1-m-m^2}=\frac{m+m^2-1}{1+m}\cdot\frac{1+m}{1-m-m^2}=\frac{m^2+m-1}{ {-}(m^2+m-1)}={-}1\)

• \({\largeг)}\ \left(a^{\backslash{c}}+\frac{a^2}{c}\right):\left(b^{\backslash{a}}+\frac{bc}{a}\right)=\frac{ac+a^2}{c}:\frac{ab+bc}{a}=\frac{ a(c+a)}{c}\cdot\frac{a}{ b(a+c)}=\frac{a}{c}\cdot\frac{a}{b}=\frac{a^2}{bc}\)

• \({\largeд)}\ \left(\frac{a+x^{\backslash{x\ -\ a}}}{x\phantom{^{\backslash{x\ -\ a}}}}-\frac{2x^{\backslash{x}}}{x-a\phantom{^{\backslash{x}}}}\right):\frac{a^2+x^2}{x-a}=\frac{ (a+x)\cdot(x-a)-2x\cdot{x}}{ x(x-a)}:\frac{a^2+x^2}{x-a}=\frac{x^2-a^2-2x^2}{ x(x-a)}\cdot\frac{x-a}{a^2+x^2}=\frac{{-}a^2-x^2}{ x(a^2+x^2)}=\frac{ {-}(a^2+x^2)}{ x(a^2+x^2)}={-}\frac{1}{x}\)

• \({\largeе)}\ \left(\frac{x^2+1^{\backslash2}}{2x-1\phantom{^{\backslash2}}}-\frac{x^{\backslash{2x\ -\ 1}}}{2\phantom{^{\backslash{2x\ -\ 1}}}}\right)\cdot\frac{1-2x}{x+2}=\frac{ (x^2+1)\cdot2-x\cdot(2x-1)}{ 2(2x-1)}\cdot\frac{1-2x}{x+2}=\frac{2x^2+2-2x^2+x}{ 2(2x-1)}\cdot\frac{ {-}(2x-1)}{x+2}=\frac{2+x}{2}\cdot\left({-}\frac{1}{2+x}\right)={-}\frac{1}{2}\)

• \({\largeж)}\ \left(\frac{n^{\backslash{n\ -\ x}}}{n+x\phantom{^{\backslash{n\ -\ x}}}}-\frac{n^{\backslash{n\ +\ x}}}{n-x\phantom{^{\backslash{n\ +\ x}}}}\right):\left(\frac{n^{\backslash{n\ +\ x}}}{n-x\phantom{^{\backslash{n\ +\ x}}}}+\frac{n^{\backslash{n\ -\ x}}}{n+x\phantom{^{\backslash{n\ -\ x}}}}\right)=\frac{ n\cdot(n-x)-n\cdot(n+x)}{ (n+x)(n-x)}:\frac{ n\cdot(n+x)+n\cdot(n-x)}{ (n+x)(n-x)}=\frac{n^2-nx-n^2-nx}{n^2-x^2}\cdot\frac{n^2-x^2}{n^2+nx+n^2-nx}=\frac{{-}2nx}{1}\cdot\frac{1}{2n^2}={-}\frac{x}{n}\)

• \({\largeз)}\ \frac{3}{5x}-\frac{3}{x+y}\cdot\left(\frac{x+y^{\backslash1}}{5x\phantom{^{\backslash1}}}-x^{\backslash5x}-y^{\backslash5x}\right)=\frac{3}{5x}-\frac{3}{x+y}\cdot\frac{ (x+y)\cdot1-x\cdot5x-y\cdot5x}{5x}=\frac{3}{5x}-\frac{3}{x+y}\cdot\frac{x+y-5x^2-5xy}{5x}=\frac{3^{\backslash{x\ +\ y}}}{5x\phantom{^{\backslash{x\ +\ y}}}}-\frac{3x+3y-15x^2-15xy^{\backslash1}}{ 5x(x+y)\phantom{^{\backslash1}}}=\frac{ 3\cdot(x+y)-(3x+3y-15x^2-15xy)\cdot1}{ 5x(x+y)}=\frac{3x+3y-3x-3y+15x^2+15xy}{ 5x(x+y)}=\frac{15x^2+15xy}{ 5x(x+y)}=\frac{ 15x(x+y)}{ 5x(x+y)}=3\)