Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
Ta có:
\(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}=\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)^2}{\left(1+\sqrt{n}+\sqrt{n+1}\right)\left(1-\sqrt{n}+\sqrt{n+1}\right)}=\frac{2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}}{2\left(1+\sqrt{n+1}\right)}\)
\(=\frac{\left[2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}\right]\left(1-\sqrt{n+1}\right)}{2\left(1+\sqrt{n+1}\right)\left(1-\sqrt{n+1}\right)}=\frac{-2n\sqrt{n+1}+2n\sqrt{n}}{-2n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra:
\(Q=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-\sqrt{2}< \sqrt{2017}-1=R\)
Vậy Q < R.
\(Tongquat:\)
\(\sqrt{1+\frac{1}{n}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n}+\frac{2}{n}-\frac{2}{n+1}-\frac{2}{n\left(n+1\right)}+\frac{1}{\left(n+1\right)^2}}\)
\(=\sqrt{\left(1+\frac{1}{n}\right)^2-2\left(1+\frac{1}{n}\right)\frac{1}{n+1}+\frac{1}{n+1}}=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}\)
\(=|1+\frac{1}{n}-\frac{1}{n+1}|=1+\frac{1}{n}-\frac{1}{n+1}\)
Thay vào ta có:
\(P=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+.........-\frac{1}{2017}\)
\(P=2015+\frac{1}{2}-\frac{1}{2017}=2015+\frac{2015}{4034}\)
a )\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\)
=\(\sqrt{2+3+1+2\sqrt{2.1+2\sqrt{3}.1+2\sqrt{2}.\sqrt{3}}}\)
=\(\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}\)
=\(\sqrt{2}+\sqrt{3}+1\)
ta có: \(\left(\sqrt{2017^2-1}-\sqrt{2016^2-1}\right)\left(\sqrt{2017^2-1}+\sqrt{2016^2-1}\right)\)
= 20172-1 - (20162-1)
= 20172-20162
= 2017+2016 > 2.2016
=> \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)\(>\) \(\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
So sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)và \(\frac{2\cdot2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Ta có :
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}=\frac{2017^2-1-2016^2+1}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\frac{2017+2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
\(>\frac{2016+2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Vậy \(\sqrt{2017^2-1}-\sqrt{2016^2-1}>\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Với mọi \(n\in N.\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}.\)Do đó
\(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}.=1-\frac{1}{\sqrt{2017}}=\frac{\sqrt{2017}-1}{\sqrt{2017}}.\)