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\(A=\left(\frac{x-1}{\sqrt{x}-1}+\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\right).\frac{1}{2\sqrt{x}}=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x-1}}+\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right].\frac{1}{2\sqrt{x}}\)
\(A=2\left(\sqrt{x}+1\right).\frac{1}{2\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}>1=\sqrt{\frac{2019}{2019}}>\sqrt{\frac{2018}{2019}}\) ( đpcm )
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ta xét : \(\sqrt{a^2+b^2+\frac{a^2}{\left(\frac{a}{b}+1\right)^2}}=\sqrt{\left(a+b\right)^2-2ab+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\left(a+b\right)^2-2.\left(a+b\right).\frac{ab}{a+b}+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\left(a+b-\frac{ab}{a+b}\right)^2}=\left|a+b-\frac{ab}{a+b}\right|\)
áp dụng vào bài toán :
\(A=\left|1+2018-\frac{2018}{2019}\right|+\frac{2018}{2019}=2019\)
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
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
\(=1-\frac{1}{\sqrt{2017}}\)
\(\frac{1}{\sqrt{k\left(2018-k+1\right)}}>\frac{2}{k+2019-k}=\frac{2}{2019}\)
Ap dụng bài toan được
\(A>\frac{2}{2019}+\frac{2}{2019}+...+\frac{2}{2019}=2.\frac{2018}{2019}\)