A=\(\frac{5}{11\times16}+\frac{5}{16\times21}+......+\frac{5}{61\times66}\)
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đặt A= dãy số trên.Ta có:
5A= \(\frac{5}{11x16}+\frac{5}{16x21}+...+\frac{5}{61x66}\)
=> 5A= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
=> 5A = \(\frac{1}{11}-\frac{1}{66}\)
=> 5A= \(\frac{5}{66}\)
=> A=\(\frac{1}{66}\)
\(=\frac{1}{5}\left(\frac{5}{11.16}\frac{5}{16.21}\frac{5}{21.26}+......+\frac{5}{61.66}\right)\)
\(=\frac{1}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}+\frac{1}{66}\right)\)
=\(\frac{1}{5}\left(\frac{1}{11}+\frac{1}{66}\right)\)
\(=\frac{1}{5}.\frac{7}{66}\)
\(=\frac{7}{330}\)
\(\frac{1}{11\times16}+\frac{1}{16\times21}+\frac{1}{21\times26}+...+\frac{1}{56\times61}+\frac{1}{61\times66}\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\frac{5}{66}\)
\(=\frac{1}{66}\)
\(\frac{1}{11\times16}+\frac{1}{16\times21}+\frac{1}{21\times26}+...+\frac{1}{56\times61}+\frac{1}{61\times66}\)
\(=\frac{1}{5}\times\left(\frac{5}{11\times16}+\frac{5}{16\times21}+\frac{5}{21\times26}+...+\frac{5}{56\times61}+\frac{5}{61\times66}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}+\frac{1}{61}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{11}-\frac{1}{66}\right)\)
\(=\frac{1}{5}\times\frac{5}{66}=\frac{1}{66}\)
Dấu " . " là dấu nhân
\(\frac{16\cdot17-5}{16\cdot16+11}\)
\(=\frac{16\cdot\left(16+1\right)-5}{16^2+11}\)
\(=\frac{16^2+16-5}{16^2+11}\)
\(=\frac{16^2+11}{16^2+11}\)
\(=1\)
Ta có: \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}.\frac{1}{120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}.\frac{1}{80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{1}{64}+\frac{5}{2003}\)
Vì \(\left\{\begin{matrix}\frac{1}{64}>\frac{1}{180}\\\frac{5}{2003}>\frac{2}{2003}\end{matrix}\right.\Rightarrow\frac{1}{64}+\frac{5}{2003}>\frac{1}{180}+\frac{2}{2003}\Rightarrow B>A\)
Vậy A < B
\(B=\frac{3+\frac{8}{12}+\frac{9}{13}-\frac{12}{17}}{4+\frac{8}{12}+\frac{12}{13}-\frac{16}{17}}+\frac{4+\frac{16}{60}-\frac{24}{213}-\frac{32}{11}}{5+\frac{20}{61}-\frac{36}{213}-\frac{40}{11}}\)
\(\Leftrightarrow B=\frac{3\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}{4\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}+\frac{4\left(1+\frac{4}{60}-\frac{6}{213}-\frac{8}{11}\right)}{5\left(1+\frac{4}{60}+\frac{6}{213}-\frac{8}{11}\right)}\)
\(\Leftrightarrow B=\frac{3}{4}+\frac{4}{5}\)
\(\Leftrightarrow B=\frac{15}{20}+\frac{16}{20}\)
\(\Leftrightarrow B=\frac{31}{20}\)
\(\frac{1}{1\times3}\) \(+\) \(\frac{1}{3\times5}\) \(+\) \(\frac{1}{5\times7}\) \(+\) \(...\) \(\frac{1}{19\times21}\)
\(=\) \(\frac{943}{1995}\)
A=1-1 phần 2 +1 phần 2 -1 phần 3.... + 1phan19 - 1 phần 21
A= 1- 1 phần 21
A= 20 phần 21
Ta có: \(D=\frac{\frac{15}{6x16}+\frac{15}{16x26}+\frac{15}{26x36}}{\frac{33}{6x16}-\frac{63}{16x26}+\frac{93}{26x36}}\)
\(\Rightarrow D=\frac{15.\frac{1}{6x16}+15.\frac{1}{16x26}+15.\frac{1}{26x36}}{3.11.\frac{1}{6x16}-3.21.\frac{1}{16x26}+3.31.\frac{1}{26x36}}\)
\(\Rightarrow D=\frac{15.\left(\frac{1}{6x16}+\frac{1}{16x26}+\frac{1}{26x36}\right)}{3.\left(\frac{11}{6x16}-\frac{21}{16x26}+\frac{31}{26x36}\right)}\)
\(\Rightarrow D=5.\left(\frac{1}{6x16}+\frac{1}{16x26}+\frac{1}{26x36}\right):\left(\frac{11}{6x16}-\frac{21}{16x26}+\frac{31}{26x36}\right)\)
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)
\(=\frac{5}{8}:\frac{15}{16}\)
\(=\frac{2}{3}\)
\(A=\frac{16-11}{11.16}+\frac{21-16}{16.21}+...+\frac{66-61}{61.66}\)
\(A=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)