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22 tháng 6 2016
\(\frac{1}{M}=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+...+\frac{1}{1+2+3+...+59}\)
\(\frac{1}{M}=\frac{1}{3\left(1+3\right):2}+\frac{1}{4\left(1+4\right):2}+\frac{1}{5\left(1+5\right):2}+...+\frac{1}{59\left(1+59\right):2}\)
\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)
\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{60}\right)\)
\(\frac{1}{M}=\frac{1}{2}.\frac{19}{60}\)
\(\frac{1}{M}=\frac{19}{120}\)
\(M=\frac{120}{19}>\frac{2}{3}\left(đpcm\right)\)
18 tháng 8 2016
Ta có : \(\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\)
\(\Rightarrow m=\frac{30}{19}>\frac{2}{3}\)
NT
18 tháng 8 2016
\(Tac\text{ó}:\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{59}-\frac{1}{60}\right)\)
\(=>2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\\ =>m=\frac{30}{19}>\frac{2}{3}\)
NT
0
TB
4
24 tháng 4 2016
Ta có:
\(1+2+3=\frac{\left(1+3\right).3}{2}=\frac{4.3}{2}\)
\(1+2+3+4=\frac{\left(1+4\right).4}{2}=\frac{5.4}{2}\)
\(.................\)
\(1+2+3+...+59=\frac{\left(1+59\right).59}{2}=\frac{60.59}{2}\)
Nên : \(M=\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+...+\frac{1}{\frac{60.59}{2}}\) , suy ra :\(2M=\frac{1}{4.3}+\frac{1}{5.4}+...+\frac{1}{60.59}\)
\(2M=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\)
\(2M=\frac{1}{3}-\frac{1}{60}<\frac{1}{3}\)
Do đó : \(M<\frac{1}{3}.2\)
\(M<\frac{2}{3}\)
1 tháng 6 2018
Bài 1 :
Ta có;\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}>\frac{1}{30}.10=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}.30>\frac{1}{30}.24=\frac{2}{5}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\left(1\right)\)
Mặt khác :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}.20=1\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}.20=\frac{1}{2}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< 1+\frac{1}{2}=\frac{3}{2}\left(2\right)\)
Từ (1 ) và (2) ta suy ra điều phải chứng minh
Bài 2 :
Đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)
MỘT MẶT ,TA CÓ THỂ VIẾT
\(S=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)\(+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}\right)-\frac{1}{64}\)
\(>\frac{1}{2}.2+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32-\frac{1}{64}\)\(=\frac{7}{2}-\frac{1}{64}=\frac{223}{64}>\frac{192}{64}=3\left(1\right)\)
Mặt khác ,ta lại có\(S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\)\(+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)\)\(+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)< \)\(1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32=6\left(2\right)\)
Từ (1) và (2 ) ta kết luận \(3< S< 6\)
Chúc bạn học tốt ( -_- )
8 tháng 8 2017
\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)
8 tháng 8 2017
B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)
= \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)
= \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)
= \(\frac{-22}{105}\)
C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}=\frac{6}{7}\)
CMR nhé
\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+59}\)
\(=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)
\(=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)
\(=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{59.60}\right)\)
\(=2.\left(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{60-59}{59.60}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{60}\right)\)
\(=2.\frac{19}{60}\)
\(=\frac{38}{60}\)\(< \frac{40}{60}=\frac{2}{3}\)