Tính tống
\(S=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}\)\(+....+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}\)
HELP!!!
Những câu hỏi liên quan
tính B=\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}\)
Tính \(C=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Ta có: \(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{k\left(k+1\right)^2-k^2\left(k+1\right)}\)
\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k^3+2k^2+k-k^3-k^2}\)
\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k\left(k+1\right)}\)
\(=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Lần lượt thay k=1;2;...;2018 ta được:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{1}-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
...
\(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}=\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
Cộng vế theo vế ta được:
\(C=1-\frac{1}{\sqrt{2019}}=...\)
https://i.imgur.com/rbOpKwh.jpg
Chứng minh :
A = \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)
là 1 số hữu tỉ .
bn có thể tham khảo ở sách vũ hữu binh nha
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Tính:
A= \(\frac{1}{2\sqrt{1}+1\sqrt{2}}\)+ \(\frac{1}{3\sqrt{2}+2\sqrt{3}}\)+....+ \(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)
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:
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
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\(t\text{ổng}qu\text{át}:\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{n^2\left(n-1\right)-\left(n-1\right)^2n}\)
\(=\frac{n\sqrt{n-1}-\left(n-1\right)\sqrt{n}}{\left(n-1\right)n}\)
\(=\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\)
Thay vào A có
\(A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}\)
\(=1-\frac{1}{\sqrt{2017}}\)
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Tính: \(B=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{2018^2}+\frac{1}{2019^2}}\)
Chứng minh: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) với a+b+c=0
Sau đó thay vào B tính ra
Số không đẹp lắm đâu 
https://i.imgur.com/BGVzR9S.jpg
Tính tổng S= \(\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{2019^2}+\sqrt{2019^2-2}}\)
\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)
\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)
\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)
\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)
\(S=\frac{\left(2019-1\right)}{2}=1009\)
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\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)
\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)
\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)
\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)
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Xem thêm câu trả lời
Rút gọn biểu thức S = \(\frac{2019}{2\sqrt{1}+1\sqrt{2}}+\frac{2019}{3\sqrt{2}+2\sqrt{3}}+\frac{2019}{4\sqrt{3}+3\sqrt{4}}+...+\frac{2019}{2019\sqrt{2018}+2018\sqrt{2019}}\)
Mk chỉ cần kết quả thôi , cảm ơn nhiều ạ
Rút gọn \(\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}+\frac{1-\sqrt{4}+\sqrt{5}}{1+\sqrt{4}+\sqrt{5}}+...+\frac{1-\sqrt{2018}+\sqrt{2019}}{1+\sqrt{2018}+\sqrt{2019}}\)
tính
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}-\sqrt{5}}+...+\frac{1}{\sqrt{2018}-\sqrt{2019}}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{2018}-\sqrt{2019}}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2019}-\sqrt{2018}}{2019-2018}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}\)
\(=\sqrt{2019}-\sqrt{2}\)
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