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Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}=\frac{49}{50}\)
B=\(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}+\frac{3}{17.20}\)
\(\Rightarrow B=\frac{5-2}{2.5}+\frac{8-5}{5.8}+...+\frac{17-14}{14.17}+\frac{20-17}{17.20}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{20}=\frac{10}{20}-\frac{1}{20}=\frac{9}{20}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)
c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{10}{11}\)
\(=\frac{1.2.3.....10}{2.3.4.....11}=\frac{1}{11}\)
Bài 1
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ... + \(\frac{1}{99.100}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{99}\) - \(\frac{1}{100}\)
= 1 - \(\frac{1}{100}\)
= \(\frac{99}{100}\)
Còn những bài kia em không biết làm vì em mới học lớp 6.
Chúc anh/chị học tốt!
Bài 1
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Bài 3:
b)\(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\)
Ta thấy: \(\begin{cases}\left|2x-27\right|^{2011}\ge0\\\left(3y+10\right)^{2012}\ge0\end{cases}\)
\(\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)
\(\Rightarrow\begin{cases}\left|2x-27\right|^{2011}=0\\\left(3y+10\right)^{2012}=0\end{cases}\)\(\Rightarrow\begin{cases}2x-27=0\\3y+10=0\end{cases}\)\(\Rightarrow\begin{cases}2x=27\\3y=-10\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}\)
a) = 2(1-1/2+1/2-1/3+...+1/19-1/20)
= 2(1-1/20)
= 2.19/20
= 19/10
b) = 7(1/2-1/3+1/3-1/4+...+1/6-1/7)
= 7(1/2 - 1/7)
= 7.5/14
= 5/2
c) = 1/2-1/5+1/5-1/8+...+1/14-1/17
= 1/2 - 1/17
= 15/34
Chúc bạn học tốt nhé
a)2/1.2+2/2.3+....+2/19.20
=2(1/1.2+1/2.3+....+1/19.20)
=2(1-1/2+1/2-1/3+.....-1/20)
=2(1-1/20)
2(19/20)=38/20=19/10
b)7/2.3+7/3.4+7/4.5+7/5.6+7/6.7
7(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7)
7(1/2-1/3+1/3-1/4+.....-1/7)
7(1/2-1/7)
7(7/14-2/14)=7.5/14=35/14=5/2
c)3/2.5+3/5.8+3/8.11+3/11.14+3/14.17
1/2-1/5+1/5-1/8+......+1/14-1/17
1/2-1/17=17/34-2/34=15/34