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Với mọi \(n\inℕ^∗\) ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n-1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\)
Áp dụng đẳng thức trên ta có:
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
\(t\text{ổng}qu\text{át}:\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{n^2\left(n-1\right)-\left(n-1\right)^2n}\)
\(=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{\left(n-1\right)n}\)
\(=\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\)
Thay vào A có
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
\(=1-\frac{1}{\sqrt{2017}}\)
\(A=\left(\frac{x-1}{\sqrt{x}-1}+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}}=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x-1}}+\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right].\frac{1}{2\sqrt{x}}\)
\(A=2\left(\sqrt{x}+1\right).\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}>1=\sqrt{\frac{2019}{2019}}>\sqrt{\frac{2018}{2019}}\) ( đpcm )
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@Thế Vĩ@
\(P=\sqrt{2}.\frac{\sqrt{2020}-\sqrt{2}}{2}=\sqrt{2}.\frac{\sqrt{2}\left(\sqrt{1010}-1\right)}{2}=2.\frac{\sqrt{1010}-1}{2}=\sqrt{1010}-1\)
Áp dụng \(\frac{1}{\sqrt{a.b}}>\frac{2}{a+b}\) , ta có :
\(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+...+\frac{1}{\sqrt{k\left(1998-k+1\right)}}+...+\frac{1}{\sqrt{1998.1}}>\)
\(>\frac{2}{1+1998}+\frac{2}{2+1997}+...+\frac{2}{k+1998-k+1}+...+\frac{2}{1998+1}=\)
\(=\frac{2.1998}{1999}\)
Vậy \(S>\frac{2.1998}{1999}\)
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:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)
\(\frac{1}{\sqrt{k\left(2018-k+1\right)}}>\frac{2}{k+2019-k}=\frac{2}{2019}\)
Ap dụng bài toan được
\(A>\frac{2}{2019}+\frac{2}{2019}+...+\frac{2}{2019}=2.\frac{2018}{2019}\)