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A, 1/(1.2)+...+1/[x(x+1)]=99
=>1-1/2+...+1/x+1/(x-1)=99
=>1-1/(x-1)=99
=>1/(x-1)=-98
=>1/(x-1)=-98/1
=>1.(-98)=(x-1).1(tích chéo)
=>x-1=-98
=>x=-97
a. \(\frac{1}{1.2}+...+\frac{1}{x.\left(x+1\right)}=99\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+...+\frac{1}{x}-\frac{1}{x+1}=99\)
\(\Rightarrow1-\frac{1}{x+1}=99\)
\(\Rightarrow\frac{1}{x+1}=1-99=-98\)
\(\Rightarrow x=\frac{1}{-98}-1\)
\(\Rightarrow x=-\frac{99}{98}\)
P/s : Bạn ơi đề sai, x sai hay mk sai ạ???
\(\left|x-3\right|=2x+4\)
\(\left|x-3\right|=2x+2\cdot2\)
\(\left|x-3\right|=2\left(x+2\right)\)
\(\Rightarrow\orbr{\begin{cases}x-3=-\left[2\cdot\left(x+2\right)\right]\\x-3=2\left(x+2\right)\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-3=-\left[2x+4\right]\\x-3=2x+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=-2x-4\\x=2x+2+3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-2x-4+3\\x=2x+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2x-1\\x=2x+5\end{cases}}\) \(.....................\)
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
Em tham khảo nhé!
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}=\frac{2009}{4018}-\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
=> \(1\cdot2009=1\left(x+1\right)\)
=> \(x+1=2009\Rightarrow x=2009-1=2008\)
Vậy x = 2008
Chúc bn hk tốt !
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
đặt A = 1/1.3 + 1/3.5 + ..+ 1/19.21
2A = 2/1.3 + 2/3.5 +..+2/19.21
2A = 3-1/1.3 +...+5-3/3.5 +... + 21-19/19.21
2A = 1/1 - 1/3 + 1/3 -1/5 +...+1/19 -1/21
2A = 20/21
A = 10/21
r thế vào tính là xng
\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)