Tính \(S=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\)\(\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
Tính: \(P=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{24\sqrt{25}+25\sqrt{24}}\)
Bạn viết sai phân số cuối cùng.
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2\sqrt{1}-1\sqrt{2}}{\left(2\sqrt{1}+1\sqrt{2}\right)\left(2\sqrt{1}-1\sqrt{2}\right)}=\frac{2\sqrt{1}-1\sqrt{2}}{\left(2\sqrt{1}\right)^2-\left(1\sqrt{2}\right)^2}=\frac{2\sqrt{1}-1\sqrt{2}}{2^21-1^22}=\frac{2\sqrt{1}-1\sqrt{2}}{1.2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Tương tự:
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{3\sqrt{2}-2\sqrt{3}}{3^22-2^23}=\frac{3\sqrt{2}-2\sqrt{3}}{2.3}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
....
\(\frac{1}{25\sqrt{24}+24\sqrt{25}}=\frac{25\sqrt{24}-24\sqrt{25}}{25^224-24^225}=\frac{25\sqrt{24}-24\sqrt{25}}{25.24}=\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
Vậy \(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
rút gọn:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
CMR:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}< 1\)
\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)
\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)
Tính
P=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}\)+\(\frac{1}{3\sqrt{2}+2\sqrt{3}}\)+.................+\(\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
Chị tham khảo ở đây ạ:
Câu hỏi của Vũ Thảo Vy - Toán lớp 9 - Học toán với OnlineMath
CMR
\(\sqrt{n}< \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+........+\frac{1}{\sqrt{n}}< 2\sqrt{n}\)
\(\sqrt{1}+\sqrt{2}+.......+\sqrt{n}< n\sqrt{\frac{n+1}{2}}\)
\(\frac{\sqrt{2}-\sqrt{1}}{1+2}+\frac{\sqrt{3}-\sqrt{2}}{2+3}+.........+\frac{\sqrt{25}-\sqrt{24}}{24+25}< \frac{2}{5}\)
a. Chứng minh : \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. Áp dụng : Tính giá trị của biểu thức :
\(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
cảm ơn các bạn trước nhé!
\(A=\frac{\sqrt{2}-\sqrt{1}}{2+1}+\frac{\sqrt{3}-\sqrt{2}}{3+2}+...+\frac{\sqrt{25}+\sqrt{24}}{25+24}\)
CMR A<0,4
Tính A=\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
Lời giải :
Xét dạng tổng quát sau :
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Từ đó ta có hướng giải quyết bài toán :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}\)
\(A=\sqrt{25}-\sqrt{1}\)
\(A=4\)
Tính giá trị biểu thức:
\(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
Có \(\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\left(n\ge2\right)=\frac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}=\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}=\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\)
=> \(\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}=\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\)(1)
Áp dụng (1) vào bt M có:
M=\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)=\(1-\frac{1}{5}\)=\(\frac{4}{5}\)
Vậy M=\(\frac{4}{5}\)