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1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1......
\(\Leftrightarrow\)2/26+2/28+2/30+.....+2/50=1-1/2+1/3-1......
\(\Leftrightarrow\)1/13+1/14+1/15+.....+1/25=1-1/2+1/3-1......
\(\Leftrightarrow\)2/14+2/16+2/18+.....+2/24=1-1/2+1/3-1......
\(\Leftrightarrow\)1/7+1/8+1/9+........+1/12=1-1/2+1/3-1/4+......
\(\Leftrightarrow\)2/8+2/10+2/12+....=1-1/2+1/3-1/4+1/5-1/6
\(\Leftrightarrow\)1/4+1/5+1/6=1-1/2+1/3-1/4+1/5-1/6
\(\Leftrightarrow\)2/4+2/6=1-1/2+1/3
\(\Leftrightarrow\)1/2+1/3=1-1/2+1/3
\(\Leftrightarrow\)2/2=1
Đề sai đúng không đáng lẽ phải như này
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\left(đpcm\right)\)
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{40}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)
\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)
\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)
\(=\frac{1}{30}\cdot\frac{31}{2}\)
\(=\frac{31}{60}\)
b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Ta có:
\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)
\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)
\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)
\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)
Mà \(\frac{3}{2}< 2\)
\(\Rightarrow1< A< 2\)
c ,Ta có
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 7/14
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 <1/14 +1/14 +1/14 +1/14 +1/14 +1/14 +1/14
dù 1/3>1/14 nhưng :1/30<1/14 1/32<1/14 ;1/35<1/14 ;1/45<1/14 ;1/47<1/14 ;1/50<1/14
nên: 1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 1/2
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 7/14
1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 <1/14 +1/14 +1/14 +1/14 +1/14 +1/14 +1/14
dù 1/3>1/14 nhưng :1/30<1/14 1/32<1/14 ;1/35<1/14 ;1/45<1/14 ;1/47<1/14 ;1/50<1/14
nên: 1/3+1/30+1/32+1/35+1/45 +1/47 +1/50 < 1/2
Ta có: \(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
\(\Rightarrow P=\left(1+\frac{1}{49}\right)+\left(1+\frac{2}{48}\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+1\)
\(\Rightarrow P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(\Rightarrow P=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)}=\frac{1}{50}\)
Vậy \(\frac{S}{P}=\frac{1}{50}\)
(1/21+1/22+...+1/30)+(1/31+...+1/40)+(1/41+...+1/50)
(1/21+1/22+...+1/30)<1/20+..+1/20=1/20*10=1/2
(1/31+...+1/40)<1/30+..+1/30=1/30*10=1/3
(1/41+...+1/50)<1/40+...+1/40=1/40*10=1/4
Suy ra day so <1/2+1/3+1/4=13/12=1/1/12=>dpcm
k cho minh nhe
Vế trái=\(\left(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{50}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)=Vế phải
Vậy vế trái = vế phải (đpcm)