CMR : \(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}< 18\)
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BS
1
15 tháng 10 2017
ta có :
\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}\)
\(=\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+...+\dfrac{2}{2\sqrt{100}}\)
\(>\dfrac{2}{\sqrt{1}+\sqrt{2}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}+\dfrac{2}{\sqrt{3}+\sqrt{4}}+...+\dfrac{2}{\sqrt{100}+\sqrt{101}}\)
\(=2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{100}+\sqrt{101}}\right)\)
\(=2\left(\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}+...+\dfrac{\sqrt{100}-\sqrt{101}}{100-101}\right)\)
\(=2\left(\dfrac{\sqrt{1}-\sqrt{101}}{-1}\right)=2\left(\sqrt{101}-\sqrt{1}\right)=18,1\)
\(>18\)
\(\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>18\)
1 tháng 8 2018
Ta có : \(\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}< \dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}\left(1\right)\)
\(\sqrt{n}-\sqrt{n-1}=\dfrac{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n+1}\right)}{\sqrt{n}+\sqrt{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}>\dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)>\dfrac{1}{\sqrt{n}}\left(2\right)\)
Từ \(\left(1;2\right)\text{⇒ }đpcm\)
nè
KV
1
TM
5 tháng 7 2017
Xét dạng tổng quát:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\frac{1}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\)
\(< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Áp dụng vào bài toán:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)
\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)
\(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)
\(< 2\left(1-\frac{1}{\sqrt{2004}}\right)\)
\(< 2-\frac{2}{\sqrt{2004}}< 2\)
=>đpcm
AT
11 tháng 6 2021
\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)
11 tháng 6 2021
cả 2 ý bạn trục căn thức ở mấu là xong nhé:
vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy
11 tháng 6 2021
Với n\(\in N\)* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)
a) Áp dụng (*) vào T
\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)
\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)
Vậy n=24.
22 tháng 12 2023
a: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\dfrac{3}{2}\sqrt{2}+\dfrac{1}{2}\sqrt{2}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)
\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)
b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{18}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)
c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
d: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=-\sqrt[3]{27}=-3\)
e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}=0\)
Đặt A=\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}\)
\(\Leftrightarrow A=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+....+\dfrac{2}{2\sqrt{100}}\)
\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+....+\dfrac{2}{\sqrt{99}+\sqrt{99}}+\dfrac{2}{\sqrt{100}+\sqrt{100}}\)
\(\Leftrightarrow A=2\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{99}}+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\)
Ta có:
\(\dfrac{1}{\sqrt{2}+\sqrt{2}}< \dfrac{1}{1+\sqrt{2}};\dfrac{1}{\sqrt{3}+\sqrt{3}}< \dfrac{1}{\sqrt{2}+\sqrt{3}}\)
Tường tự, ta có:
\(\dfrac{A}{2}< \dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(A< 2\left(\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\right)\)
\(A< -2\left(1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...-\sqrt{99}+\sqrt{99}-\sqrt{100}\right)\)
\(A< -2\left(1-\sqrt{100}\right)\)
\(A< 18\)
Vậy\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}< 18\)