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S
4 tháng 5 2023
\(A=2017:\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2018}\right)\)
\(=2017:\dfrac{2017}{2018}\)
\(=2017\cdot\dfrac{2018}{2017}\)
\(=2018\)
#NgDat
NK
4 tháng 5 2023
\(A=2017:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+...+\dfrac{1}{2017}\cdot\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{2018}{2018}-\dfrac{1}{2018}\right)\)
\(A=2017:\dfrac{2017}{2018}\)
\(A=2018.\)
TN
3
21 tháng 2 2023
Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.
\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)
Ta có:
\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)
\(S=1-\dfrac{1}{2018}\)
\(S=\dfrac{2017}{2018}\)
21 tháng 2 2023
=1/1.2+1/2.3+1/3.4+...1/2017.2018
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018
=1-1/2018
=2018/2018-1/2018
=2017/2018
31 tháng 12 2017
\(\frac{1}{1.2}\)\(+\frac{1}{2.3}+\)\(\frac{1}{3.4}\)\(+\)\(.............+\)\(\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
31 tháng 12 2017
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{2018-2017}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
a) \(A=\frac{1}{5}-\frac{1}{5^2}+\frac{1}{5^3}-\frac{1}{5^4}+...+\frac{1}{5^{99}}-\frac{1}{5^{100}}\)
\(\Rightarrow5A=1-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^3}+...+\frac{1}{5^{98}}-\frac{1}{5^{99}}\)
\(\Rightarrow5A+A=1-\frac{1}{5^{100}}\)
\(A=\frac{1-\frac{1}{5^{100}}}{6}\)
b) B = 1.2+2.3+3.4+...+2017.2018
=>3B=1.2.3 + 2.3.3+3.4.3+...+2017.2018.3
3B = 1.2.3 + 2.3.(4-1) +3.4.(5-2) +...+2017.2018.(2019-2016)
3B = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2017.2018.2019-2016.2017.2018
3B = 2017.2018.2019
\(B=\frac{2017.2018.2019}{3}\)
3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2017.2018.3
3B = 1.2.3 + 2.3.(4-1) + 3.4.(5-2)+...+ 2017.2018(2019-2016)
3B = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018
3B = 2017.2018.2019
B = 2017.2018.2019/3
B= 2739315938