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\(\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{\left(n-1\right)^2-1}{\left(n-1\right)^2}\right)\left(\frac{n^2-1}{n^2}\right)\)
=\(\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}.\frac{\left(4-1\right)\left(4+1\right)}{4^2}...\frac{\left(n-2\right)n}{\left(n-1\right)^2}.\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)
=\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{\left(n-2\right).n}{\left(n-1\right)^2}.\frac{\left(n-1\right)\left(n+1\right)}{n^2}=\frac{1}{2}.\frac{n+1}{n}=\frac{1}{2}+\frac{1}{2n}>\frac{1}{2}\)
Xét dạng tổng quát có: \(\frac{1}{\sqrt{n+1}\left(n+1\right)+n\sqrt{n}}=\frac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left[n-\sqrt{n\left(n+1\right)}+n+1\right]}=\frac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)-\sqrt{n\left(n+1\right)}}\)
\(< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}-\sqrt{n\left(n+1\right)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta có:
\(\frac{1}{2\sqrt{2}+1\sqrt{1}}< 1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{3}+2\sqrt{2}}< \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
.....
\(\frac{1}{\left(n+1\right)\sqrt{n+1}+n\sqrt{n}}< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cộng vế theo vế =>\(VT< 1-\frac{1}{\sqrt{n+1}}\left(ĐPCM\right)\)
ta có \(\left(1+\frac{1}{k}-\frac{1}{k-1}\right)^2\)
= \(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)\(+\frac{2}{k-1}-\frac{2}{k}-\frac{2}{k\left(k-1\right)}\)
=\(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2k-2k+2-2}{k\left(k-1\right)}\)
= \(1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)
=> \(\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}\)= \(1+\frac{1}{k-1}-\frac{1}{k}\)(đpcm)
à mà thôi khỏi giải t biết làm rồi