cmr: $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+…+\frac{1}{\sqrt{97}+\sqrt{99}}>\frac{9}{4}$ 03/10/2021 Bởi Bella cmr: $\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+…+\frac{1}{\sqrt{97}+\sqrt{99}}>\frac{9}{4}$
$\dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}} > \dfrac {9}{4}$ + Đặt: $A = \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}}$. $⇔ 2A = \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}}$. $⇔ 2A > \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {3} + \sqrt {5}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}}$. $⇔ 2A > -\dfrac {\sqrt {1} – \sqrt {3} + \sqrt {3} – \sqrt {5} + … + \sqrt {97} – \sqrt {99}}{2}$. $⇔ 2A > \dfrac {\sqrt {99} – 1}{2}$. $⇔ A > \dfrac {\sqrt {99} – 1}{4} > \dfrac {9}{4}$. + Vậy: $A > \dfrac {9}{4}$. ——————— XIN HAY NHẤT CHÚC EM HỌC TỐT Bình luận
$\dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}} > \dfrac {9}{4}$
+ Đặt: $A = \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}}$.
$⇔ 2A = \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}}$.
$⇔ 2A > \dfrac {1}{\sqrt {1} + \sqrt {3}} + \dfrac {1}{\sqrt {3} + \sqrt {5}} + \dfrac {1}{\sqrt {5} + \sqrt {7}} + … + \dfrac {1}{\sqrt {97} + \sqrt {99}}$.
$⇔ 2A > -\dfrac {\sqrt {1} – \sqrt {3} + \sqrt {3} – \sqrt {5} + … + \sqrt {97} – \sqrt {99}}{2}$.
$⇔ 2A > \dfrac {\sqrt {99} – 1}{2}$.
$⇔ A > \dfrac {\sqrt {99} – 1}{4} > \dfrac {9}{4}$.
+ Vậy: $A > \dfrac {9}{4}$.
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XIN HAY NHẤT
CHÚC EM HỌC TỐT