Question

Problema 5 va rog dau coroana

Answer 1

Răspuns:

sume telescopice.

Explicație pas cu pas:

1/[n(n+1)] = 1/n - 1/(n+1) (se verifică prin aducere la același numitor)

inecuația devine

1/2-1/3+1/3-1/4+1/4-1/5+....+1/(2x)-1/(2x+1)<3/5

rămâne 1/2-1/(2x+1)<3/5

De aici cred că te descurci.

Answer 2

Explicație pas cu pas:

$x \neq0 \ ; \ x \not= - \dfrac{1}{2}$

$\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + ... + \dfrac{1}{2x \cdot (2x + 1)} < \dfrac{3}{5}$

$\dfrac{2 - 1}{1 \cdot 2} + \dfrac{3 - 2}{2 \cdot 3} + \dfrac{4 - 3}{3 \cdot 4} + ... + \dfrac{2x + 1 - 2x}{2x \cdot (2x + 1)} < \dfrac{3}{5}$

$\Big(\dfrac{1}{1} - \dfrac{1}{2}\Big) + \Big(\dfrac{1}{2} - \dfrac{1}{3}\Big) + \Big(\dfrac{1}{3} - \dfrac{1}{4}\Big) + ... + \Big(\dfrac{1}{2x} - \dfrac{1}{2x + 1}\Big) < \dfrac{3}{5}$

$\dfrac{1}{1} - \dfrac{1}{2x + 1} < \dfrac{3}{5}$

$\dfrac{2}{5} - \dfrac{1}{2x + 1} < 0$

$\dfrac{4x - 3}{5(2x + 1)} < 0$

$4x - 3 = 0 \implies x = - \dfrac{3}{4}$

$\begin{cases} x < - \dfrac{3}{4} \implies 4x - 3 < 0\\x \geqslant - \dfrac{3}{4} \implies 4x - 3 \geqslant 0 \end{cases}$

$2x + 1 = 0 \implies x = - \dfrac{1}{2}$

$\begin{cases} x < - \dfrac{1}{2} \implies 2x + 1 < 0\\x > - \dfrac{1}{2} \implies 2x + 1 > 0 \end{cases}$

$\implies x \in \Big(- \dfrac{1}{2} ; \dfrac{3}{4} \Big) - \Big\{ 0\Big\}$