第83页

信息发布者：
解：​$ x + 2y - 1 = 0，$​则​$x + 2y = 1。$​
故原式​$=\frac {2(x + 2y)}{(x + 2y)^2} = \frac {2}{x + 2y}。$​
将​$x + 2y = 1$​代入，得​$\frac {2}{1} = 2。$​
解：因为​$\frac {a}{a + b} = k，$​所以​$a = k(a + b)；$​
同理​$\frac {c}{c + d} = k$​得​$c = k(c + d)，$​​$\frac {e}{e + f} = k$​得​$e = k(e + f)。$​
则​$a + c + e = k(a + b) + k(c + d) + k(e + f) $​
​$= k[(a + b) + (c + d) + (e + f)]。$​
因此​$\frac {a + c + e}{a + b + c + d + e + f} = \frac {k[(a + b) + (c + d) + (e + f)]}{(a + b) + (c + d) + (e + f)} = k。$​
上一页 下一页