Упражнение 235

\({\largeа)}\ \frac{5}{y-3}+\frac{1}{y+3}-\frac{4y-18}{y^2-9}=\frac{5}{y-3}+\frac{1}{y+3}-\frac{4y-18}{ (y-3)(y+3)}=\frac{ 5(y+3)+(y-3)-(4y-18)}{ (y-3)(y+3)}=\frac{5y+15+y-3-4y+18}{ (y-3)(y+3)}=\frac{2y+30}{ (y-3)(y+3)}=\frac{2y+30}{y^2-9};\)

\({\largeб)}\ \frac{2a}{2a+3}+\frac{5}{3-2a}-\frac{4a^2+9}{4a^2-9}=\frac{2a}{2a+3}-\frac{5}{2a-3}-\frac{4a^2+9}{ (2a-3)(2a+3)}=\frac{ 2a(2a-3)-5(2a+3)-(4a^2+9)}{ (2a-3)(2a+3)}=\frac{4a^2-6a-10a-15-4a^2-9}{ (2a-3)(2a+3)}=\frac{{-}16a-24}{ (2a-3)(2a+3)}=\frac{{-}8(2a+3)}{ (2a-3)(2a+3)}={-}\frac{8}{2a-3}=\frac{8}{3-2a};\)

\({\largeв)}\ \frac{4m}{4m^2-1}-\frac{2m+1}{6m-3}+\frac{2m-1}{4m+2}=\frac{4m}{ (2m-1)(2m+1)}-\frac{2m+1}{ 3(2m-1)}+\frac{2m-1}{ 2(2m+1)}=\frac{6\cdot4m-2(2m+1)(2m+1)+3(2m-1)(2m-1)}{ 6(2m-1)(2m+1)}=\frac{24m-2(4m^2+4m+1)+3(4m^2-4m+1)}{ 6(2m-1)(2m+1)}=\frac{24m-8m^2-8m-2+12m^2-12m+3}{ 6(2m-1)(2m+1)}=\frac{4m^2+4m+1}{ 6(2m-1)(2m+1)}=\frac{ (2m+1)^2}{ 6(2m-1)(2m+1)}=\frac{2m+1}{ 6(2m-1)}=\frac{2m+1}{12m-6};\)

\({\largeг)}\ \frac{1}{ (x+y)^2}-\frac{2}{x^2-y^2}+\frac{1}{ (x-y)^2}=\frac{1}{ (x+y)^2}-\frac{2}{ (x-y)(x+y)}+\frac{1}{ (x-y)^2}=\frac{ (x-y)^2-2(x-y)(x+y)+(x+y)^2}{ (x-y)^2(x+y)^2}=\frac{x^2-2xy+y^2-2x^2+2y^2+x^2+2xy+y^2}{ (x-y)^2(x+y)^2}=\frac{4y^2}{ (x-y)^2(x+y)^2};\)

\({\largeд)}\ \frac{4a^2+3a+2}{a^3-1}-\frac{1-2a}{a^2+a+1}=\frac{4a^2+3a+2}{ (a-1)(a^2+a+1)}-\frac{1-2a}{a^2+a+1}=\frac{4a^2+3a+2-(1-2a)(a-1)}{ (a-1)(a^2+a+1)}=\frac{4a^2+3a+2-(a-1-2a^2+2a)}{ (a-1)(a^2+a+1)}=\frac{4a^2+3a+2-a+1+2a^2-2a}{ (a-1)(a^2+a+1)}=\frac{6a^2+3}{ (a-1)(a^2+a+1)}=\frac{6a^2+3}{a^3-1};\)

\({\largeе)}\ \frac{x-y}{x^2+xy+y^2}-\frac{3xy}{x^3-y^3}+\frac{1}{x-y}=\frac{x-y}{x^2+xy+y^2}-\frac{3xy}{ (x-y)(x^2+xy+y^2)}+\frac{1}{x-y}=\frac{ (x-y)(x-y)-3xy+x^2+xy+y^2}{ (x-y)(x^2+xy+y^2)}=\frac{x^2-2xy+y^2-3xy+x^2+xy+y^2}{ (x-y)(x^2+xy+y^2)}=\frac{2x^2-4xy+2y^2}{ (x-y)(x^2+xy+y^2)}=\frac{ 2(x^2-2xy+y^2)}{ (x-y)(x^2+xy+y^2)}=\frac{ 2(x-y)^2}{ (x-y)(x^2+xy+y^2)}=\frac{ 2(x-y)}{x^2+xy+y^2}=\frac{2x-2y}{x^2+xy+y^2}.\)