Question-132292

Question Number 132292 by Algoritm last updated on 13/Feb/21

Answered by mathmax by abdo last updated on 13/Feb/21

\frac{1}{[x]} + \frac{1}{[2 x]} = [x] + \frac{1}{3} let [x] = n \Rightarrow n ⩽ x < n + 1 \Rightarrow 2 n ⩽ 2 x < 2 n + 2

if 2 x \in [2 n, 2 n + 1 [\Rightarrow x \in [n, n + \frac{1}{2} [\Rightarrow [2 x] = 2 n

e \Rightarrow \frac{1}{n} + \frac{1}{2 n} = n + \frac{1}{3} \Rightarrow \frac{3}{2 n} = n + \frac{1}{3} \Rightarrow \frac{9}{2 n} = 3 n + 1 \Rightarrow 9 = 2 n (3 n + 1) \Rightarrow

{6 n}^{2} + 2 n - 9 = 0 \to Δ^{^{'}} = 1 - 6 . (- 9) = 1 + 54 = 55 \Rightarrow n_{1} = \frac{- 1 + \sqrt{54}}{12}

not solution also n_{2} = \frac{- 1 - \sqrt{54}}{12} not solution (n integr!)

if 2 x \in [2 n + 1, 2 n + 2 [\Rightarrow x \in [n + \frac{1}{2}, n + 1 [[2 x] = 2 n + 1

e \Rightarrow \frac{1}{n} + \frac{1}{2 n + 1} = n + \frac{1}{3} \Rightarrow \frac{3 n + 1}{n (2 n + 1)} = \frac{3 n + 1}{3} \Rightarrow

{2 n}^{2} + n = 3 \Rightarrow {2 n}^{2} + n - 3 = 0

Δ = 1 - 4 (2) (- 3) = 1 + 24 = 25 \Rightarrow n_{1} = \frac{- 1 + 5}{4} = 1 and n_{2} = \frac{- 1 - 5}{4} = - \frac{3}{2}

not solution \Rightarrow n = 1 \Rightarrow x \in [\frac{3}{2}, 2 [this is the set of solution