\({\largeа)}\ \frac{2y^2-y}{y^2-y+\frac{1}{4}}-\frac{2y^2+y}{y^2+y+\frac{1}{4}}-\frac{1}{y^2-\frac{1}{4}}=\frac{ 4(2y^2-y)}{ 4\left(y^2-y+\frac{1}{4}\right)}-\frac{ 4(2y^2+y)}{ 4\left(y^2+y+\frac{1}{4}\right)}-\frac{4}{ 4\left(y^2-\frac{1}{4}\right)}=\frac{ 4y(2y-1)}{4y^2-4y+1}-\frac{ 4y(2y+1)}{4y^2+4y+1}-\frac{4}{4y^2-1}=\frac{ 4y(2y-1)}{ (2y-1)^2}-\frac{ 4y(2y+1)}{ (2y+1)^2}-\frac{4}{ (2y-1)(2y+1)}=\frac{4y}{2y-1}-\frac{4y}{2y+1}-\frac{4}{ (2y-1)(2y+1)}=\frac{ 4y(2y+1)-4y(2y-1)-4}{ (2y-1)(2y+1)}=\frac{8y^2+4y-8y^2+4y-4}{ (2y-1)(2y+1)}=\frac{8y-4}{ (2y-1)(2y+1)}=\frac{ 4(2y-1)}{ (2y-1)(2y+1)}=\frac{4}{2y+1};\)
\({\largeб)}\ \frac{6a}{2{,}25a^2-0{,}64}-\frac{8}{6a-3{,}2}=\frac{6a}{ (1{,}5a-0{,}8)(1{,}5a+0{,}8)}-\frac{8}{ 4(1{,}5a-0{,}8)}=\frac{6a\cdot4-8(1{,}5a+0{,}8)}{ 4(1{,}5a-0{,}8)(1{,}5a+0{,}8)}=\frac{24a-12a-6{,}4}{ 4(1{,}5a-0{,}8)(1{,}5a+0{,}8)}=\frac{12a-6{,}4}{ 4(1{,}5a-0{,}8)(1{,}5a+0{,}8)}=\frac{ 8(1{,}5a-0{,}8)}{ 4(1{,}5a-0{,}8)(1{,}5a+0{,}8)}=\frac{2}{1{,}5a+0{,}8}=\frac{2\cdot10}{ (1{,}5a+0{,}8)\cdot10}=\frac{20}{15a+8}.\)