Ответ:
- а) Снимаем внешние скобки: \(-\frac12x+\frac23y-\left(\frac32x-\frac14\right)=-\frac12x+\frac23y-\frac32x+\frac14=-2x+\frac23y+\frac14\).
- б) \(\frac85\left(\frac52s-\frac58\right)-\frac23\left(12s+\frac32\right)=4s-1-8s-1=-4s-2\).
- в) Знаменатель общий — 6: \(\frac{4y-3}{3}-\frac{y-5}{2}=\frac{8y-6-3y+15}{6}=\frac{5y+9}{6}\).
- г) \(\frac{3(4p+5)-2(1+3p)}5=\frac{12p+15-2-6p}{5}=\frac{6p+13}{5}\).
- д) \(\frac{7x-4}{2}-\frac{12x-9}{6}-\frac{4(x+5)}8=\frac{21x-12-(12x-9)-3(x+5)}6=\frac{6x-18}{6}=x-3\).
- е) \(\frac{2(2p-q)}2-\frac{3(q-4p)}2+\frac14=2p-q-\frac32q+6p+\frac14=8p-\frac52q+\frac14\).
Ответ: а) \(-2x+\frac23y+\frac14\); б) \(-4s-2\); в) \(\frac{5y+9}{6}\); г) \(\frac{6p+13}{5}\); д) \(x-3\); е) \(8p-\frac52q+\frac14\).
