1/12+1/24+1/40+...+1/1200
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PN
1
21 tháng 2 2021
Xong rồi bạn ơi ko có người đâu tôi cũng thế đây này chán nhể
3 tháng 10 2016
b) Áp dụng t/c dãy tỉ số bằng nhau,ta có:
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+3y+1+5y+1+7y}{12+5x+4x}=\frac{3+15y}{12+5x+4x}=\frac{3\left(1+5y\right)}{2.3.2+5x+4x}=\frac{1+5y}{4+9x}=\frac{1+5y}{5x}\)<=> 4 + 9x = 5x
....
4 tháng 10 2016
a/ Từ giả thiêt ta có \(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\Leftrightarrow\frac{x}{15}-\frac{3}{5}=\frac{y}{20}-\frac{3}{5}=\frac{z}{40}-\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\). Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{40}=k\)
\(\Rightarrow\begin{cases}x=15k\\y=20k\\z=40k\end{cases}\)
Theo đề bài : \(xy=1200\Leftrightarrow15k.20k=1200\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Tới đây dễ rồi nhé :)
b/ \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\Leftrightarrow\frac{1+5y}{5}=\frac{1+7y}{4}\Leftrightarrow\frac{7+35y}{35}=\frac{5+35y}{20}=\frac{7+35y-5-35y}{35-20}=\frac{2}{15}\)
\(\Rightarrow y=-\frac{1}{15}\)
Thay y vào \(\frac{1+3y}{12}=\frac{1+5y}{5x}\) tìm được x = 2
3 tháng 8 2015
2A=1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
14 tháng 8 2016
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)
\(=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)
14 tháng 8 2016
=\(\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+...+\frac{2}{9800}\)
=\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{98.100}\)
=\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
2 tháng 7 2023
=1/2(1/2+1/6+1/12+1/20+...+1/90)
=1/2(1-1/2+1/2-1/3+...+1/9-1/10)
=1/2*9/10=9/20
BD
11 tháng 7 2023
\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}+x=1\)
\(\)\(2\cdot\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}\right):2+x=1\)
\(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right):2+x=1\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right):2+x=1\)
\(\dfrac{6}{7}:2+x=1\)
\(x=1-\dfrac{3}{7}\)
\(x=\dfrac{4}{7}\)
\(\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 7 2023
\(\dfrac{1}{4}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{40}\) + \(\dfrac{1}{60}\) + \(\dfrac{1}{84}\) + \(x\) = 1
\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.6}\)+\(\dfrac{1}{2.12}\)+\(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.30}\) + \(\dfrac{1}{2.42}\) + \(x\) =1
\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\) + \(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).\(\dfrac{6}{7}\) + \(x\) = 1
\(\dfrac{3}{7}\) + \(x\) = 1
\(x\) = 1 - \(\dfrac{3}{7}\)
\(x\) = \(\dfrac{4}{7}\)
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