Упражнение 285

\({\largeа)}\ \left(1-\frac{3x^2}{1-x^2}\right):\left(\frac{x}{x+1}+1\right)=\frac{1-x^2-3x^2}{1-x^2}:\frac{x+x+1}{x+1}=\frac{1-4x^2}{1-x^2}:\frac{2x+1}{x+1}=\frac{ (1-2x)(1+2x)}{ (1-x)(1+x)}\cdot\frac{1+x}{1+2x}=\frac{1-2x}{1-x};\)

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

\({\largeв)}\ \frac{3a^2-a+3}{a^3-1}-\frac{a-1}{a^2+a+1}+\frac{2}{1-a}=\frac{3a^2-a+3}{ (a-1)(a^2+a+1)}-\frac{a-1}{a^2+a+1}-\frac{2}{a-1}=\frac{3a^2-a+3-(a-1)^2-2(a^2+a+1)}{ (a-1)(a^2+a+1)}=\frac{3a^2-a+3-(a^2-2a+1)-2a^2-2a-2}{ (a-1)(a^2+a+1)}=\frac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{ (a-1)(a^2+a+1)}=\frac{{-}a}{ (a-1)(a^2+a+1)}={-}\frac{a}{a^3-1};\)

\({\largeг)}\ \left(\frac{2b}{1-b}-b\right):\frac{3b+3}{b-1}=\frac{2b-b(1-b)}{1-b}\cdot\frac{b-1}{3b+3}=\frac{2b-b+b^2}{1-b}\cdot\frac{b-1}{ 3(b+1)}=\frac{b^2+b}{1-b}\cdot\frac{b-1}{ 3(b+1)}=\frac{ b(b+1)}{{-}(b-1)}\cdot\frac{b-1}{ 3(b+1)}={-}\frac{b}{3};\)

\({\largeд)}\ \left(a-x+\frac{x^2}{a+x}\right)\cdot\frac{a-x}{a}=\frac{ (a-x)(a+x)+x^2}{a+x}\cdot\frac{a-x}{a}=\frac{a^2-x^2+x^2}{a+x}\cdot\frac{a-x}{a}=\frac{a^2}{a+x}\cdot\frac{a-x}{a}=\frac{ a(a-x)}{a+x}=\frac{a^2-ax}{a+x}.\)