Упражнение 99

\({\largeа)}\ \frac{2a+b}{2a^2-ab}-\frac{16a}{4a^2-b^2}-\frac{2a-b}{2a^2+ab}=\frac{2a+b}{ a(2a-b)}-\frac{16a}{ (2a-b)(2a+b)}-\frac{2a-b}{ a(2a+b)}=\frac{ (2a+b)(2a+b)-16a\cdot{a}-(2a-b)(2a-b)}{ a(2a-b)(2a+b)}=\frac{4a^2+4ab+b^2-16a^2-(4a^2-4ab+b^2)}{ a(2a-b)(2a+b)}=\frac{4a^2+4ab+b^2-16a^2-4a^2+4ab-b^2}{ a(2a-b)(2a+b)}=\frac{{-}16a^2+8ab}{ a(2a-b)(2a+b)}=\frac{{-}8a(2a-b)}{ a(2a-b)(2a+b)}={-}\frac{8}{2a+b};\)

\({\largeб)}\ \frac{1}{ (a-3)^2}-\frac{2}{a^2-9}+\frac{1}{ (a+3)^2}=\frac{1}{ (a-3)^2}-\frac{2}{ (a-3)(a+3)}+\frac{1}{ (a+3)^2}=\frac{ (a+3)^2-2(a-3)(a+3)+(a-3)^2}{ (a-3)^2(a+3)^2}=\frac{a^2+6a+9-2a^2+18+a^2-6a+9}{ (a-3)^2(a+3)^2}=\frac{36}{ (a-3)^2(a+3)^2}=\frac{36}{ (a^2-9)^2};\)

\({\largeв)}\ \frac{x-2}{x^2+2x+4}-\frac{6x}{x^3-8}+\frac{1}{x-2}=\frac{x-2}{x^2+2x+4}-\frac{6x}{ (x-2)(x^2+2x+4)}+\frac{1}{x-2}=\frac{ (x-2)(x-2)-6x+x^2+2x+4}{ (x-2)(x^2+2x+4)}=\frac{x^2-4x+4-6x+x^2+2x+4}{ (x-2)(x^2+2x+4)}=\frac{2x^2-8x+8}{ (x-2)(x^2+2x+4)}=\frac{ 2(x^2-4x+4)}{ (x-2)(x^2+2x+4)}=\frac{ 2(x-2)^2}{ (x-2)(x^2+2x+4)}=\frac{ 2(x-2)}{x^2+2x+4};\)

\({\largeг)}\ \frac{2a^2+7a+3}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{3}{a-1}=\frac{2a^2+7a+3}{ (a-1)(a^2+a+1)}-\frac{1-2a}{a^2+a+1}-\frac{3}{a-1}=\frac{2a^2+7a+3-(1-2a)(a-1)-3(a^2+a+1)}{ (a-1)(a^2+a+1)}=\frac{2a^2+7a+3-(a-1-2a^2+2a)-3a^2-3a-3}{ (a-1)(a^2+a+1)}=\frac{2a^2+7a+3-a+1+2a^2-2a-3a^2-3a-3}{ (a-1)(a^2+a+1)}=\frac{a^2+a+1}{ (a-1)(a^2+a+1)}=\frac{1}{a-1}.\)