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14 tháng 4 2019

\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)

\(\Rightarrow T=\frac{1004}{1005}\)

14 tháng 4 2019

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\frac{2010}{2011}\)

\(\Rightarrow A=\frac{1005}{2011}\)

14 tháng 4 2019

a) C=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)

\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}\)\(=\frac{1.2.3...999}{2.3.4...1000}=\frac{1.\left(2.3.4....999\right)}{\left(2.3.4....999\right).1000}\)\(=\frac{1}{1000}\)

b) Đặt: A=\(1+2+2^2+2^3+...+2^{2008}\)

\(\Leftrightarrow2A=2+2^2+2^3+....+2^{2008}+2^{2009}\)

\(\Leftrightarrow2A-A=2^{2009}-1\)

\(\Leftrightarrow A=2^{2009}-1\)

\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)\(=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=\frac{1}{-1}=-1\)

vậy: S=(-1)

NV
16 tháng 5 2019

\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)

\(\frac{-17}{2.4}-\frac{17}{4.6}-\frac{17}{6.8}-...-\frac{17}{100.102}\)

\(=-\frac{17}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)

\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\right)\)

\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{102}\right)\)

\(=-\frac{17}{2}.\frac{25}{51}=-\frac{25}{6}\)

8 tháng 3 2019

Mk ko biết lm nhưng cứ k thoải mái nha

SORRY

22 tháng 11 2021

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)

12 tháng 4 2016

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19 tháng 3 2019

Bài 2:

a) \(\frac{4}{9}+x=\frac{-5}{3}\)

\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)

\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)

Vậy: \(x=\frac{-19}{9}\)

b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)

\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)

\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)

c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)

\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-5\right\}\)