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第89页

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解：$(2)$

$\begin {aligned}∵\frac {5x}{(x + 1)(x - 3)}&=\frac {a}{x + 1}+\frac {b}{x - 3}\\∴去分母，得5x&=(a + b)x+(b - 3a)\\∴\begin {cases}a + b = 5\\b - 3a = 0\end {cases}\\解得\begin {cases}a=\frac {5}{4}\\b =\frac {15}{4}\end {cases}\end {aligned}$

$(3)$

$\begin {aligned}&\frac {1}{(x + 1)(x + 2)}+\frac {1}{(x + 2)(x + 3)}+\frac {1}{(x + 3)(x + 4)}+···+\frac {1}{(2x + 1)(2x + 2)}\\=&\frac {1}{x + 1}-\frac {1}{x + 2}+\frac {1}{x + 2}-\frac {1}{x + 3}+\frac {1}{x + 3}-\frac {1}{x + 4}+···+\frac {1}{2x + 1}-\frac {1}{2x + 2}\\=&\frac {1}{x + 1}-\frac {1}{2x + 2}\\=&\frac {1}{2x + 2}\end {aligned}$