Найдите такие значения a и b, при которых выполняется тождество:
а) $\frac{5x}{(x - 2)(x + 3)} = \frac{a}{x - 2} + \frac{b}{x + 3}$;
б) $\frac{5x + 31}{(x - 5)(x + 2)} = \frac{a}{x - 5} - \frac{b}{x + 2}$.
$\frac{5x}{(x - 2)(x + 3)} = \frac{a}{x - 2} + \frac{b}{x + 3}$
$\frac{5x}{(x - 2)(x + 3)} = \frac{a(x + 3) + b(x - 2)}{(x - 2)(x + 3)} = \frac{ax + 3a + bx - 2b}{(x - 2)(x + 3)} = \frac{5x}{(x - 2)(x + 3)} = \frac{x(a + b) + 3a - 2b}{(x - 2)(x + 3)}$
\begin{equation*}
\begin{cases}
a + b = 5 &\\
3a - 2b = 0 &
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
2a + 2b = 10 &\\
3a - 2b = 0 &
\end{cases}
\end{equation*}
2a + 2b + 3a − 2b = 10 + 0
5a = 10
a = 2
b = 5 − a = 5 − 2 = 3
Ответ: а = 2; b = 3.
$\frac{5x + 31}{(x - 5)(x + 2)} = \frac{a}{x - 5} - \frac{b}{x + 2}$
$\frac{5x + 31}{(x - 5)(x + 2)} = \frac{a(x + 2) - b(x - 5)}{(x - 5)(x + 2)} = \frac{ax + 2a - bx + 5b}{(x - 5)(x + 2)} = \frac{5x + 31}{(x - 5)(x + 2)} = \frac{x(a - b) + 2a + 5b}{(x - 5)(x + 2)}$
\begin{equation*}
\begin{cases}
a - b = 5 &\\
2a + 5b = 31 &
\end{cases}
\end{equation*}
\begin{equation*}
\begin{cases}
-2a + 2b = -10 &\\
2a + 5b = 31 &
\end{cases}
\end{equation*}
−2a + 2b + 2a + 5b = −10 + 31
7b = 21
b = 3
a = 5 + b = 8
Ответ: а = 8; b = 3.
Пожауйста, оцените решение
