Cho B = \(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+ ... + \(\dfrac{1}{99.100}\)
Chứng minh \(\dfrac{7}{12}\)<B<\(\dfrac{5}{6}\)
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24 tháng 7 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+..+\dfrac{1}{9900}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9900}\right)\)
\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< \dfrac{5}{6}\left(2\right)\)
\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm\)
24 tháng 7 2017
Ta có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{99.100}\)
\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+............+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}\)
\(\Leftrightarrow A>\dfrac{1}{12}\)\(\left(1\right)\)
Lại có :
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...........+\dfrac{1}{99.100}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-.........-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow A< \dfrac{5}{6}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm\)
VD
20 tháng 3 2022
d, `3,15+2,4=5,55`
e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)
f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`
\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
20 tháng 3 2022
\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)
\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)
12 tháng 7 2017
2, a-b=ab => a=ab+b => a=b(a+1)
thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1
Theo bài ra ta có: a:b=a-b
=> a+1=a-b
=>-b=1
=> b=-1
Thay b=-1 vào a-b=ab ta có : a-(-1)=-a
=> a +1=-a
=>a=-1/2
Vậy a=-1/2. b=-1
HD
14 tháng 11 2017
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\right)\)
\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< \dfrac{5}{6}\left(2\right)\)
\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(\rightarrowđpcm\)
Chúc bạn học tốt!
18 tháng 9 2021
\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\\ =\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{49}+\dfrac{1}{50}\)
MC
5 tháng 7 2018
Ta có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\Rightarrow A>\dfrac{1}{1.2}+\dfrac{1}{3.4}=\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{14}{24}=\dfrac{7}{12}\)\(\left(1\right)\)
Lại có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-\left(\dfrac{1}{6}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)\(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
Vậy \(\dfrac{7}{12}< A< \dfrac{5}{6}\) ( Điều phải chứng minh ).
5 tháng 7 2018
Ta có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}\right)+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\dfrac{7}{12}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}>\dfrac{7}{12}\left(1\right)\\ \Rightarrow A>\dfrac{7}{12}\left(1\right)\)
Ta lại có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\\ A=\dfrac{5}{6}-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}< \dfrac{5}{6}\\ \Rightarrow A=< \dfrac{5}{6}\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{7}{12}< A< \dfrac{5}{6}\left(dpcm\right)\)
20 tháng 6 2017
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
16 tháng 4 2017
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
11 tháng 2 2022
a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\).
b. Có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\).
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}< 1\)
Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{247}{300}\)
Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)
Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)
\(\Rightarrowđpcm\)