Cho mk hỏi nek!Có thưởng ak nha:
Tính:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\cdot2+2\cdot3+3\cdot4+...+98\cdot99}\)
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QD
19 tháng 2 2017
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
28 tháng 1 2019
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
ML
27 tháng 6 2015
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Rightarrow k=2\)
3 tháng 4 2019
Tính tổng dãy dấu ngoặc trước
Đặt \(S=1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot(4-1)+...+98\cdot99\cdot(100-97)\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot3\cdot4+...+98\cdot99\cdot100-97\cdot98\cdot99\)
\(3S=98\cdot99\cdot100\Rightarrow S=\frac{1}{3}\cdot98\cdot99\cdot100\)
Thay vào đề bài,ta có :
\(\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}:\frac{-3}{2}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}\cdot\frac{2}{-3}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{90}{7}\cdot\frac{2}{-3}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-30}{7}\cdot\frac{2}{-1}\)
\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-60}{-7}=\frac{60}{7}\)
\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=\frac{60}{7}\cdot26950\)
\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=231000\)
\(\Leftrightarrow323400\cdot x=231000\)
\(\Leftrightarrow x=231000:323400=\frac{5}{7}\)
7 tháng 4 2019
Tử thần sai từ dòng:
\(\frac{\frac{1}{3}.98.99.100.x}{26950}=\frac{30}{7}.\frac{2}{-1}\Leftrightarrow12x=-\frac{60}{7}\Leftrightarrow x=\frac{-5}{7}\)
26 tháng 6 2019
\(\left(1\cdot2\right)^{-1}+\left(2\cdot3\right)^{-1}+\cdot\cdot\cdot+\left(9\cdot10\right)^{-1}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
26 tháng 6 2019
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(\frac{\frac{1.2}{2}+\frac{2.3}{2}+...+\frac{98.99}{2}}{1.2+2.3+3.4+...+98.99}=\frac{1}{2}\)
Ta gọi:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+97\cdot98}\) là A
\(=\frac{1+3+6+10+...+4753}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
\(\Rightarrow2A=\frac{2+6+12+20+...+9506}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
\(=\frac{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}{1\cdot2+2\cdot3+3\cdot4+...+97\cdot98}\)
=> 2A = 1
=> A = 1/2
Mik nghĩ bỏ 98.98 ở phần MS thì bài mới đúng, bn nên hỏi lại thầy bn
Vậy A = 1/2