tính
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}\)
Những câu hỏi liên quan
Chứng minh :\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{2018\sqrt{2017}}< 2\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đây ta có
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\)
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Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}\)
\(\Leftrightarrow\sqrt{n}\left(\frac{1}{n}-\frac{1}{n1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\). Mà:
\(\left(\frac{\sqrt{n}}{\sqrt{n+1}}+1\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Từ đó, ta có:
\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2018}}\right)< 2\) (ĐPCM)
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Rút gọn : \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{4}-\sqrt{3}\right)}+...+\frac{\sqrt{2018}-\sqrt{2017}}{\left(\sqrt{2017}+\sqrt{2018}\right)\left(\sqrt{2018}-\sqrt{2017}\right)}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2018}-\sqrt{2017}}{2018-2017}\)
\(=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}+...+\frac{\sqrt{2018}-\sqrt{2017}}{1}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2018}-\sqrt{2017}=\sqrt{2018}-1\)
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\(=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2018}\)
\(=-\left(\sqrt{1}+\sqrt{2018}\right)\)
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Ta có :
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\)\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2018}-\sqrt{2017}\)
\(=\)\(\sqrt{2018}-1\)
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Xem thêm câu trả lời
Chứng minh rằng \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}}< 2018\)
Rút gọn: \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.........+\frac{1}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{2-1}{\sqrt{1}+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+........+\frac{2018-2017}{\sqrt{2017}+\sqrt{2018}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}+......+\)
\(\frac{\left(\sqrt{2018}-\sqrt{2017}\right)\left(\sqrt{2018}+\sqrt{2017}\right)}{\sqrt{2017}+\sqrt{2018}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+........+\left(\sqrt{2018}-\sqrt{2017}\right)\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+......+\sqrt{2018}-\sqrt{2017}\)
\(=-\sqrt{1}+\sqrt{2018}=\sqrt{2018}-\sqrt{1}\)
Chứng minh \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}}< 2\)
Chứng minh
\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2017^2}+\frac{1}{2018^2}< 2018}\)
Gọi vế trái BPT là A.
Xét biểu thức tổng quát:
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{\left[n\left(n+1\right)\right]^2}}\\ =\frac{\sqrt{n^2\left(n^2+2n+1\right)+n^2+2n+1+n^2}}{n\left(n+1\right)}\\ =\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n\left(n+1\right)}\\ =\frac{\sqrt{\left(n^2+n+1\right)^2}}{n\left(n+1\right)}\\ =\frac{n^2+n+1}{n\left(n+1\right)}\\ =\frac{n\left(n+1\right)+n+1-n}{n\left(n+1\right)}\\ =1+\frac{1}{n}-\frac{1}{n+1}\)
Suy ra:
\(A=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\right)\) (2018 số hạng 1)
\(=2018+\frac{1}{2}-\frac{1}{2018}< 2018\)
Vậy \(A< 2018\left(đpcm\right)\).
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Bài 1: Rút gọn biểu thức:Afrac{a^3-3a+left(a^2-1right)sqrt{a^2-4}-2}{a^3-3a+left(a^2-1right)sqrt{a^2-4}+2}left(a2right)Bsqrt{frac{1}{a^2+b^2}+frac{1}{left(a+bright)^2}+sqrt{frac{1}{a^4}+frac{1}{b^4}+frac{1}{left(a^2+b^2right)^2}}}left(abne0right)Bài 2: Tính giá trị của biểu thức:Efrac{1}{1sqrt{2}+2sqrt{1}}+frac{1}{2sqrt{3}+3sqrt{2}}+frac{1}{3sqrt{4}+4sqrt{3}}+...+frac{1}{2017sqrt{2018}+2018sqrt{2017}}Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyênAleft(sqrt{57}+3sqrt{6}+sqrt{38}...
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Bài 1: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
Bài 2: Tính giá trị của biểu thức:
\(E=\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 3: Chứng minh rằng các biểu thức sau có gúa trị là số nguyên
\(A=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
Bài 1: Tính giá trị của biểu thức:frac{1}{1sqrt{2}+2sqrt{1}}+frac{1}{2sqrt{3}+3sqrt{2}}+frac{1}{3sqrt{4}+4sqrt{3}}+...+frac{1}{2017sqrt{2018}+2018sqrt{2017}}Bài 2: Chứng minh rằng các biểu thức sau có giá trị là số nguyênA left(sqrt{57}+3sqrt{6}+sqrt{38}+6right)left(sqrt{57}-3sqrt{6}-sqrt{38}+6right)B frac{2sqrt{3+sqrt{5-sqrt{13+sqrt{48}}}}}{sqrt{6}+sqrt{2}}
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Bài 1: Tính giá trị của biểu thức:\(\frac{1}{1\sqrt{2}+2\sqrt{1}}+\frac{1}{2\sqrt{3}+3\sqrt{2}}+\frac{1}{3\sqrt{4}+4\sqrt{3}}+...+\frac{1}{2017\sqrt{2018}+2018\sqrt{2017}}\)
Bài 2: Chứng minh rằng các biểu thức sau có giá trị là số nguyên
A = \(\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
B = \(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
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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