Toán CM: 1/(2√1 +1√2) + 1/(3√2+2√3)+…+1/(25√24+24√25)=4/5 13/08/2021 By Sadie CM: 1/(2√1 +1√2) + 1/(3√2+2√3)+…+1/(25√24+24√25)=4/5
Giải thích các bước giải: Ta có : $\begin{split}\dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}&=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{((n+1)\sqrt{n}+n\sqrt{n+1})((n+1)\sqrt{n}-n\sqrt{n+1})}\\&=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)^2-n^2(n+1)}\\&=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}\\&=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\end{split}$ $\to A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+..+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}$ $\to A=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+..+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}$ $\to A=1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}$ Trả lời
Giải thích các bước giải:
Ta có :
$\begin{split}\dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}&=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{((n+1)\sqrt{n}+n\sqrt{n+1})((n+1)\sqrt{n}-n\sqrt{n+1})}\\&=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)^2-n^2(n+1)}\\&=\dfrac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}\\&=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\end{split}$
$\to A=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+..+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}$
$\to A=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+..+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}$
$\to A=1-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}$