Tính
A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.......................\frac{79}{80}\)
NT
Những câu hỏi liên quan
Tính A=\(\frac{\sqrt{3}}{2}+\frac{\sqrt{5}}{4}+\frac{\sqrt{7}}{6}+...+\frac{\sqrt{81}}{80}\)
Tính B=\(\frac{\sqrt{1}}{2}+\frac{\sqrt{3}}{4}+\frac{\sqrt{5}}{6}+...+\frac{\sqrt{79}}{80}\)
Chứng tỏ P < \(\frac{8}{9}\)
P=\(\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{4}+\frac{\sqrt{7}-\sqrt{5}}{6}+...+\frac{\sqrt{81}-\sqrt{79}}{80}\)
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.....\frac{79}{80}.CMR\)
\(A<\frac{1}{9}\)
Ta đặt B=\(\frac{2}{3}.\frac{4}{5}...\frac{80}{81}\)
Mà \(\frac{1}{2}<\frac{2}{3};\frac{3}{4}<\frac{4}{5};...;\frac{79}{80}<\frac{80}{81}\)
=>A<B
=>A2<AB=\(\frac{1}{2}.\frac{2}{3}.....\frac{80}{81}=\frac{1}{81}\)
=>A2<\(\frac{1}{81}\)
=>A<\(\sqrt{\frac{1}{81}}=\frac{1}{9}\)(đpcm)
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Bài 1 : Tính
Cho A =\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+......+\frac{1}{60}>\frac{7}{12}\)
B = \(\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{5^2}+......+\frac{ }{50^{21}}\)
CMR B >\(\frac{1}{4}\)và B < \(\frac{4}{9}\)
C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.......\frac{79}{80}< \frac{1}{9}\)
Tính
a) \(\left( {\frac{4}{5} - 1} \right):\frac{3}{5} - \frac{2}{3}.0,5\)
b) \(1 - {\left( {\frac{5}{9} - \frac{2}{3}} \right)^2}:\frac{4}{{27}}\)
c)\(\left[ {\left( {\frac{3}{8} - \frac{5}{{12}}} \right).6 + \frac{1}{3}} \right].4\)
d) \(0,8:\left\{ {0,2 - 7.\left[ {\frac{1}{6} + \left( {\frac{5}{{21}} - \frac{5}{{14}}} \right)} \right]} \right\}\)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
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Bài 1 : Cho A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{79}{80}\)
Chứng minh rằng A < \(\frac{1}{9}\)
Bài 4 : Chứng minh rằng: 1.3.5.7....19 = \(\frac{11}{2}.\frac{12}{2}.\frac{13}{2}...\frac{20}{2}\)
So sánh:frac{frac{frac{1}{2}}{frac{3}{4}}}{frac{frac{5}{6}}{frac{7}{8}}}+frac{frac{frac{8}{7}}{frac{6}{5}}}{frac{frac{4}{3}}{frac{2}{1}}} vàfrac{frac{frac{1}{2}}{frac{3}{4}}+frac{frac{8}{7}}{frac{6}{5}}}{frac{frac{5}{6}}{frac{7}{8}}+frac{frac{4}{3}}{frac{2}{1}}}và frac{frac{frac{1}{2}+frac{8}{7}}{frac{3}{4}+frac{6}{5}}}{frac{frac{5}{6}+frac{4}{3}}{frac{7}{8}+frac{2}{1}}}vàfrac{frac{frac{1+8}{2+7}}{frac{3+6}{4+5}}}{frac{5+4}{frac{6+3}{2+1}}}
Đọc tiếp
So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
A=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{79}{80}.CM:A<\frac{1}{9}\)
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{79}{80}:CM:A<\frac{1}{9}\)