A= 1/2-1/3+1/4-1/5+...+1/48-1/49
CM : 1/5 < A < 2/5
K
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HP
1
KN
17 tháng 8 2020
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}\)
\(\Rightarrow1-A-\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\)
\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{25}\)
\(\Rightarrow\frac{49}{50}-A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)\)
Ta có :
\(\frac{1}{26}< \frac{1}{25};\frac{1}{27}< \frac{1}{25};\frac{1}{28}< \frac{1}{25};\frac{1}{29}< \frac{1}{25};\frac{1}{30}< \frac{1}{25};\)
\(\frac{1}{31}< \frac{1}{30};\frac{1}{32}< \frac{1}{30};..;\frac{1}{39}< \frac{1}{30};\frac{1}{40}< \frac{1}{30};\)
\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{49}< \frac{1}{40};\frac{1}{50}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)>\frac{49}{50}-\frac{4}{5}=\frac{9}{50}>\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow A>\frac{1}{5}\)( đpcm )
HP
0
NM
21 tháng 8 2017
a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
KL
22 tháng 8 2017
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran
4 tháng 6 2017
a) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
=) \(\frac{-1}{12}< x< \frac{1}{8}\)
Vì \(\frac{-1}{12}< 0;\frac{1}{8}>0\)và \(< 1\)
mà x là số nguyên =) \(x=0\)
b) \(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
=) \(\frac{-1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=) \(-1\le x< 7\)=) \(x=\left\{-1;0;1;2...;6\right\}\)
CN
1
13 tháng 3 2018
=1−12 +13 −14 +15 −16 +...+149 −150. A =(1+13 +15 +...+149 )−(12 +14 +16 +...+150 ).
A =(1+12 +13 +14 +15 +16 +...+149 ...
.........
TP
0
TP
1
21 tháng 7 2015
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{47}-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+....+\frac{1}{25}\right)\)\(=\frac{1}{26}+...+\frac{1}{50}\)