C=5/58.11+5/8.11.14+3.6/4.5+...+5/302.305.308 CMR C<1/48
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giúp mk với
5/58.11+5/8.11.14+....+5/302.305.308
Ta gọi biểu thức đó là D
\(D=\frac{5}{2}\left[\frac{1}{5.8}-\frac{1}{8.11}+....+\frac{1}{302.305}-\frac{1}{305.308}\right]\)
\(D=\frac{5}{2}.\left[\frac{1}{5.8}-\frac{1}{305.308}\right]\)
\(D=\frac{4695}{75152}\)
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cho c=5/5.8.11+5/8.11.14+5/302.305.308
c/m : c <1/48
Cho C= \(\dfrac{5}{5.8.11}+\dfrac{5}{8.11.14}+...+\dfrac{5}{302.305.308}\). Chứng minh C<\(\dfrac{1}{48}\)
\(C=\dfrac{5}{5\cdot8\cdot11}+\dfrac{5}{8\cdot11\cdot14}+...+\dfrac{5}{302\cdot305\cdot308}\\ =\dfrac{5}{6}\cdot\left(\dfrac{6}{5\cdot8\cdot11}+\dfrac{6}{8\cdot11\cdot14}+...+\dfrac{6}{302\cdot305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{5\cdot8}-\dfrac{1}{8\cdot11}+\dfrac{1}{8\cdot11}-\dfrac{1}{11\cdot14}+...+\dfrac{1}{302\cdot305}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{40}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\dfrac{1}{40}-\dfrac{5}{6}\cdot\dfrac{1}{305\cdot308}\\ =\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}\\ \dfrac{5}{6\cdot305\cdot308}>0\Rightarrow\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}< \dfrac{1}{48}\)
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Cho B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
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Cho B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
Lời giải:
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)
\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)
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Cho C=\(\frac{5}{5.8.11}\)+ \(\frac{5}{8.11.14}\)+..............+\(\frac{5}{302.305.308}\)
Chứng minh C< \(\frac{1}{48}\)
Chứng minh :
1,C=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}.C< \frac{3}{4}\)
2,D=\(\frac{1}{5^2}+\frac{1}{9^2}+...+\frac{1}{409^2}< \frac{1}{12}\)
3,E=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}< \frac{1}{48}\)
Chứng minh rằng:
1)B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)
2)C=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)<\(\frac{1}{48}\)
3)D=\(\frac{11}{9}+\frac{18}{16}+\frac{27}{25}+...+\frac{1766}{1764}\)
\(40\frac{20}{43}< D< 40\frac{20}{21}\)
Quy đồng mẫu các phân số sau
\(\dfrac{4.5^2}{5^3.6}\) và \(\dfrac{15}{1.2.3.4.5}\)
Ta có: \(\dfrac{4.5^2}{5^3.6}=\dfrac{2.2.5^2}{5^2.5.2.3}=\dfrac{2}{15}\)
\(\dfrac{15}{1.2.3.4.5}=\dfrac{3.5}{1.2.3.4.5}=\dfrac{1}{8}\)
Quy đồng: \(\dfrac{2}{15}=\dfrac{16}{120}\) ; \(\dfrac{1}{8}=\dfrac{15}{120}\)
Vậy \(\dfrac{4.5^2}{5^3.6}=\dfrac{16}{120}\) và \(\dfrac{15}{1.2.3.4.5}=\dfrac{15}{120}\)
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