Chứng minh
\(\frac{118}{87}< \frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{49}< \frac{118}{49}\)
Những câu hỏi liên quan
Chứng minh
\(\frac{118}{78}< \frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{49}< \frac{118}{49}\)
Chứng minh
\(\frac{118}{78}< \frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{49}< \frac{118}{49}\)
Chứng minh
\(\frac{118}{78}< \frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{49}< \frac{118}{49}\)
Chứng minh rằng :
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{49}+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\) (đpcm)
*đpcm = điều phải chứng minh
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Chứng minh:
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
ta có :
\(VP=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=VT\)
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Ta có: 1 - 1/2 + 1/3 - 1/4 + ... + 1/49 - 1/50
= 1 + 1/2 + 1/3 + 1/4 + ... + 1/49 + 1/50 - 2×( 1/2 + 1/4 + ... + 1/50)
= 1 + 1/2 + 1/3 + 1/4 + ... + 1/50 - (1 + 1/2 + 1/3 + ... + 1/25)
= 1/26 + 1/27 + 1/28 + ... + 1/49 + 1/50
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Tìm x : \(\frac{6:\frac{3}{5}-1\frac{1}{16}\cdot\frac{6}{7}}{4\frac{1}{5}\cdot\frac{10}{11}+5\frac{2}{11}}\)\(-\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{5}\right)\cdot\frac{12}{49}}{3\frac{1}{3}+\frac{2}{9}}\)
Chứng minh rằng:\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{49}+\frac{1}{50}=\frac{91}{50}-\frac{97}{49}+\frac{95}{48}-\frac{93}{47}+.....+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}=1\)
\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{8}\right)-\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
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Tìm x biết :
\(\left[\frac{6:\frac{3}{5}-1\frac{1}{16}.\frac{6}{7}}{4\frac{1}{5}.\frac{10}{11}+5\frac{2}{11}}-\frac{\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{15}\right).\frac{12}{49}}{3\frac{1}{3}+\frac{2}{9}}\right].x=2\frac{23}{96}\)
\(x=\frac{903}{391}\)
Bài này sử dụng MTCT đó bạn!
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\(\frac{1}{10\cdot11}\)+\(\frac{1}{11\cdot12}\)+\(\frac{1}{12\cdot13}\)+ ......................+\(\frac{1}{49\cdot50}\)
Ta có: \(\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)
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