Tính P=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}\)+\(\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Những câu hỏi liên quan
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3+3\sqrt{4}}}+...+\frac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Tính giá trị của biểu thức .
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
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So sánh Q=\(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{3}+\sqrt{4}}{1+\sqrt{3}+\sqrt{4}}+...+\frac{1-\sqrt{2016}+\sqrt{2017}}{1+\sqrt{2016}+\sqrt{2017}}\)với R=\(\sqrt{2017}-1\)
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.
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\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2016^2}+\frac{1}{2017^2}}\)
tính tổng
Tính P = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\sqrt{1+\frac{1}{4^2}+\frac{1}{5^2}}+...+\sqrt{1+\frac{1}{2016^2}+\frac{1}{2017^2}}\)
\(Tongquat:\)
\(\sqrt{1+\frac{1}{n}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n}+\frac{2}{n}-\frac{2}{n+1}-\frac{2}{n\left(n+1\right)}+\frac{1}{\left(n+1\right)^2}}\)
\(=\sqrt{\left(1+\frac{1}{n}\right)^2-2\left(1+\frac{1}{n}\right)\frac{1}{n+1}+\frac{1}{n+1}}=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}\)
\(=|1+\frac{1}{n}-\frac{1}{n+1}|=1+\frac{1}{n}-\frac{1}{n+1}\)
Thay vào ta có:
\(P=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+.........-\frac{1}{2017}\)
\(P=2015+\frac{1}{2}-\frac{1}{2017}=2015+\frac{2015}{4034}\)
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Tính:
a. \(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\)
b. \(\sqrt{1+2016^2+\frac{2016^2}{2017^2}}+\frac{2016}{2017}\)
a )\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\)
=\(\sqrt{2+3+1+2\sqrt{2.1+2\sqrt{3}.1+2\sqrt{2}.\sqrt{3}}}\)
=\(\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}\)
=\(\sqrt{2}+\sqrt{3}+1\)
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Chứng minh :
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{2017\sqrt{2016}}< 2\)< 2
so sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\) và\(\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
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
=> \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)\(>\) \(\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
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Xem thêm câu trả lời
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 \(\sqrt{2017^2-1}-\sqrt{2016^2-1}>\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
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so sánh
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)và \(\frac{2\cdot2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)