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b) \(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot...\cdot\frac{100^2}{100\cdot101}=\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{1\cdot2\cdot3\cdot4\cdot...\cdot100}\cdot\frac{\left(1\cdot2\cdot3\cdot...\cdot100\right)}{2\cdot3\cdot4\cdot...\cdot101}=1\cdot\frac{1}{101}=\frac{1}{101}\)
a không biết
Gần đúng được không
\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)cách làm \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) tách hết ra bạn thấy cái giữa tự triệt tiêu nhau
\(B=\frac{5}{1.3}+...+\frac{5}{23.25}\) { nếu đúng là \(\frac{5}{23.5}\) thì làm đến \(\frac{5}{21.23}\) rồi cộng lẻ cái cuối
\(\frac{2B}{5}=\frac{2}{1.3}+...+\frac{1}{23.25}=1-\frac{1}{25}\) cách làm giống (a)
\(\frac{2}{5}B=1-\frac{1}{25}=\frac{24}{25}\Rightarrow B=\frac{24}{25}.\frac{5}{2}=\frac{12}{5}\)
A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
=\(\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{19}-\frac{1}{20}\right)\)
=\(\frac{1}{2}-\frac{1}{20}=\frac{10-1}{20}=\frac{9}{20}\)
\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
\(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100
= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )
= 5 . ( 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 )
= 5 . ( 1 - 1/100 )
= 5 . 99/100
= 99/20
1.
a. \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
b. \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{4}{2}.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}\)
\(=\frac{200}{101}\)
A = \(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+... + \(\frac{1}{99}\)-\(\frac{1}{100}\)
A = \(\frac{1}{1}\)-\(\frac{1}{100}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây
A = \(\frac{99}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
Ta có : \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{x\left(x+1\right)}=\frac{99}{20}\)
\(\Rightarrow5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{99}{20}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{20}.\frac{1}{5}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x + 1 = 100
=> x = 99
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
<=> A=\(5\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2021\cdot2022}\right)\) \(A=5\left(1-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\right)\) \(A=5\left(1-\frac{1}{2022}\right)\)
\(A=5\cdot\frac{2021}{2022}\)
\(A=\frac{10105}{2022}\)
:0
\(\frac{5}{1.2}\) +\(\frac{5}{2\cdot3}\) +\(\frac{5}{3\cdot4}\) +......+\(\frac{5}{2021.2022}\)
= \(5\) . \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\cdots+\frac{1}{2021.2022}\right)\)
= \(5\frac{}{}.\left(1-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\right)\)
=\(5.\left(1-\frac{1}{2022}\right)\)
=\(5.\left(\frac{2022}{2022}-\frac{1}{2022}\right)\)
=\(5.\frac{2021}{2022}\)
=\(\frac{10105}{2022}\)
A= 5/1.2 + 5/2.3 + 5/3.4 + ... + 5/2021.2022
5A= 5.(1/1.2 + 1/2.3 + 1/3.4 +...+ 1/2021.2022)
A= 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/2021.2022
A= 1-1/2 + 1/2 - 1/3+1/3 - 1/4 +...+ 1/2021 - 1/2022
A= 1 - 1/2022
A= 2021/2022
Đặt $A = \dfrac{5}{1.2} + \dfrac{5}{2.3} + \dfrac{5}{3.4} + \dots + \dfrac{5}{2021.2022}$
Ta có: $A = 5 . \left( \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \dots + \dfrac{1}{2021.2022} \right)$
$= 5 . \left( 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dots + \dfrac{1}{2021} - \dfrac{1}{2022} \right)$
$= 5 . \left( 1 - \dfrac{1}{2022} \right)$
$= 5 . \dfrac{2021}{2022}$
$= \dfrac{10105}{2022}$
Vậy kết quả của phép tính là $\dfrac{10105}{2022}$.