2. Выполните действия: 1) (x − 3)/(3x + 6) − (x − 6)/(x + 2); 2) (m + 1)/(5m − 10) + (3 − m)/(4m − 8); 3) (y + 6)/(y − 6) − (y + 2)/(y + 6); 4) 3x/(4x − 4) + 5/(7 − 7x); 5) 2b/(2b + c) − 4b²/(4b² + 4bc + c²); 6) 2/(a² − 9) − 1/(a² + 3a).

Question

Вопрос:

Accepted Answer

\(\frac{x-3}{3(x+2)}-\frac{x-6}{x+2}=\frac{x-3-3(x-6)}{3(x+2)}=\frac{15-2x}{3(x+2)}\).
\(\frac{m+1}{5(m-2)}+\frac{3-m}{4(m-2)}=\frac{4(m+1)+5(3-m)}{20(m-2)}=\frac{19-m}{20(m-2)}\).
\(\frac{(y+6)^2-(y+2)(y-6)}{(y-6)(y+6)}=\frac{16y+48}{y^2-36}=\frac{16(y+3)}{y^2-36}\).
\(\frac{3x}{4(x-1)}-\frac{5}{7(x-1)}=\frac{21x-20}{28(x-1)}\).
Так как \(4b^2+4bc+c^2=(2b+c)^2\), получаем \(\frac{2b(2b+c)-4b^2}{(2b+c)^2}=\frac{2bc}{(2b+c)^2}\).
\(\frac{2}{(a-3)(a+3)}-\frac{1}{a(a+3)}=\frac{2a-(a-3)}{a(a-3)(a+3)}=\frac{1}{a(a-3)}\).

Ответ: