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\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+....+\frac{3}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\)
\(\Rightarrow\frac{1}{3}.S=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(\Rightarrow\frac{1}{3}.S=\frac{1}{1}-\frac{1}{2016}\)
\(\Rightarrow\frac{1}{3}.S=\frac{2015}{2016}\)
\(\Rightarrow S=\frac{2015}{672}\)
Vậy: \(\Rightarrow S=\frac{2015}{672}\)
Bạn giải giúp mk câu mk đăng tầm 5 phút nha!
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 = \(1-\frac{1}{100}=\frac{99}{100}\)
ai mak chẳng bt đó lak dấu nhân bn lần sau khỏi phải gt mất công
3A= 1.2.3+2.3.3+3.4.3+...........+2010.2011.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+.........+2010.2011.(2012-2009)
=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.....+2010.2011.2012-2009.2010.2011
=>3A=2010.2011.2012
=>3A=3.670.2011.2012
=>A=670.2011.2012
=>A= .......lấy máy tính mà tính
BÀI 1:
\(N=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{17.20}\)
\(N=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(N=2.\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(N=2.\frac{9}{20}\)
\(N=\frac{9}{10}\)
BÀI 2:
\(C=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3B=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3B=1.2\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(3B=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+...+98.99.100\right)\)
\(3B=99.100.101\)
\(3B=999900\)
\(\Rightarrow B=999900:3\)
\(B=333300\)
CHÚC BN HỌC TỐT!!!!
E = 1.2+2.3+3.4+......+99.100
Gấp E lên 3 lần ta có:
E . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
E . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
E . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100 E . 3 = 99.100.101
E = 99.100.101 : 3
E = 33.100.101
E = 333 300
k mik nha
E = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
=> 3E = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) +...+ 99.100.(101-98)
=> 3E = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
=> 3E = 99.100.101
=> E = 333300
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\dfrac{12}{25}x=1\)
\(\Rightarrow x=\dfrac{25}{12}\)
Vậy \(x=\dfrac{25}{12}\)
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right).x=1\)
Ta có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(=\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}\)
\(\Rightarrow\dfrac{12}{25}.x=1\Rightarrow x=1:\dfrac{12}{25}=\dfrac{25}{12}=2\dfrac{1}{12}\)
Vậy \(x=\dfrac{25}{12}\) hay \(x=2\dfrac{1}{12}\)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 2022.2023
⇒ 3B = 1.2.3 + 2.3.3 + 3.4.3. + ... + 2022.2023.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2022.2023.(2024 - 2021)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 + ... - 2021.2022.2023 + 2022.2023.2024
= 2022.2023.2024
⇒ B = 2022.2023.2024 : 3 = 2759728048
⇒ A = 2759728048/(2022 + 2023 + 2024)
= 1364176/3
Đặt B = 1.2 + 2.3 + 3.4 + ... + 2022.2023
⇒ 3B = 1.2.3 + 2.3.3 + 3.4.3. + ... + 2022.2023.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2022.2023.(2024 - 2021)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 + ... - 2021.2022.2023 + 2022.2023.2024
= 2022.2023.2024
⇒ B = 2022.2023.2024 : 3 = 2759728048
⇒ A = 2759728048/(2022 + 2023 + 2024)
= 1364176/3