Упражнение 156

\({\largeа)}\ \left(\frac{1}{y}+\frac{2}{x-y}\right)\left(x-\frac{x^2+y^2}{x+y}\right)=\frac{x-y+2y}{ y(x-y)}\cdot\frac{ x(x+y)-(x^2+y^2)}{x+y}=\frac{x+y}{ y(x-y)}\cdot\frac{x^2+xy-x^2-y^2}{x+y}=\frac{ (x+y)\cdot(xy-y^2)}{ y(x-y)\cdot(x+y)}=\frac{ (x+y)\cdot{y}(x-y)}{ y(x-y)\cdot(x+y)}=1;\)

\({\largeб)}\ \left(a+b-\frac{2ab}{a+b}\right):\left(\frac{a-b}{a+b}+\frac{b}{a}\right)=\frac{ (a+b)(a+b)-2ab}{a+b}:\frac{ a(a-b)+b(a+b)}{ a(a+b)}=\frac{a^2+2ab+b^2-2ab}{a+b}:\frac{a^2-ab+ab+b^2}{ a(a+b)}=\frac{a^2+b^2}{a+b}:\frac{a^2+b^2}{ a(a+b)}=\frac{a^2+b^2}{a+b}\cdot\frac{ a(a+b)}{a^2+b^2}=\frac{ (a^2+b^2)\cdot{a}(a+b)}{ (a+b)\cdot(a^2+b^2)}=a;\)

\({\largeв)}\ (x^2-1)\left(\frac{1}{x-1}-\frac{1}{x+1}+1\right)=\frac{x^2-1}{1}\cdot\frac{x+1-(x-1)+(x-1)(x+1)}{ (x-1)(x+1)}=\frac{ (x-1)(x+1)}{1}\cdot\frac{x+1-x+1+x^2-1}{ (x-1)(x+1)}=\frac{ (x-1)(x+1)\cdot(2+x^2-1)}{ (x-1)(x+1)}=x^2+1;\)

\({\largeг)}\ \left(m+1-\frac{1}{1-m}\right):\left(m-\frac{m^2}{m-1}\right)=\left(m+1+\frac{1}{m-1}\right):\left(m-\frac{m^2}{m-1}\right)=\frac{ (m+1)(m-1)+1}{m-1}:\frac{ m(m-1)-m^2}{m-1}=\frac{m^2-1+1}{m-1}:\frac{m^2-m-m^2}{m-1}=\frac{m^2}{m-1}:\frac{{-}m}{m-1}=\frac{m^2}{m-1}\cdot\left({-}\frac{m-1}{m}\right)={-}\frac{m^2\cdot(m-1)}{ (m-1)\cdot{m}}={-}m.\)