Chứng minh rằng: \(\sqrt{2 \sqrt[3]{3\sqrt[4]{4...\sqrt[2018]{2018}}}} < 2\)
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Những câu hỏi liên quan
\(\text{Chứng minh rằng:}2017< \sqrt{\frac{2}{1}}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+...+\sqrt[2018]{\frac{2018}{2017}}< 2018\)
Bài 1: Tính giá trị của biểu thức:frac{1}{1sqrt{2}+2sqrt{1}}+frac{1}{2sqrt{3}+3sqrt{2}}+frac{1}{3sqrt{4}+4sqrt{3}}+...+frac{1}{2017sqrt{2018}+2018sqrt{2017}}Bài 2: Chứng minh rằng các biểu thức sau có giá trị là số nguyênA left(sqrt{57}+3sqrt{6}+sqrt{38}+6right)left(sqrt{57}-3sqrt{6}-sqrt{38}+6right)B frac{2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}}{sqrt{6}+sqrt{2}}
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Bài 1: Tính giá trị của biểu thức:\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 2: Chứng minh rằng các biểu thức sau có giá trị là số nguyên
A = \(\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
B = \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Chứng minh rằng : \(2019/ \sqrt[2]{2018} + 2018/\sqrt[2]{2019} > \sqrt[2]{2018} + \sqrt[2]{2019}\)
\(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}\ge\frac{\left(\sqrt{2019}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2019}}=\sqrt{2018}+\sqrt{2019}\)
Dấu "=" ko xảy ra nên \(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}>\sqrt{2018}+\sqrt{2019}\)
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Chứng minh \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{2018\sqrt{2017}}< 2\)
Chứng minh :\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{2018\sqrt{2017}}< 2\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đây ta có
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\)
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Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(\Leftrightarrow\sqrt{n}\left(\frac{1}{n}-\frac{1}{n1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\). Mà:
\(\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đó, ta có:
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\) (ĐPCM)
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Chứng minh \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}}< 2\)
Chứng minh rằng
a) Với mọi số nguyên dương n có \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+..+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
b) \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}< \sqrt{2017}+\sqrt{2018}\)
Hộ mình vs
Câu b đề sai nha, bây giờ đặt \(a=\sqrt{2017},b=\sqrt{2018}\)
Ta có \(\frac{a^2}{b}+\frac{b^2}{a}< a+b\Leftrightarrow ab\left(\frac{a^2}{b}+\frac{b^2}{a}\right)< ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3< ab\left(a+b\right)\)(1)
Mà \(ab\left(a+b\right)\le\left(a^2-ab+b^2\right)\left(a+b\right)=a^3+b^3\)(2)
Từ (1), (2) => Sai
a) Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{k+1-k}{\left(k+1\right)\sqrt{k}}=\frac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}\)\(< \frac{2\sqrt{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}\sqrt{k}}=\frac{2}{\sqrt{k}}-\frac{2}{\sqrt{k+1}}\)
Cho k=1,2,....,n rồi cộng từng vế ta có:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)
Chứng minh rằng \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}< 2018\)
Bài 1: Rút gọn biểu thức sau:
a. \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}\)
b. \(B=\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2018\sqrt{2017}+2017\sqrt{2018}}\)
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
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a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
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b/ \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow B=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2017}}-\dfrac{1}{\sqrt{2018}}=1-\dfrac{1}{\sqrt{2018}}\)
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