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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
\(\left|x-3\right|=2x+4\)
\(\left|x-3\right|=2x+2\cdot2\)
\(\left|x-3\right|=2\left(x+2\right)\)
\(\Rightarrow\orbr{\begin{cases}x-3=-\left[2\cdot\left(x+2\right)\right]\\x-3=2\left(x+2\right)\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-3=-\left[2x+4\right]\\x-3=2x+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=-2x-4\\x=2x+2+3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-2x-4+3\\x=2x+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2x-1\\x=2x+5\end{cases}}\) \(.....................\)
a) \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)
= \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\)
= \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
= \(\frac{1}{2}.\frac{4}{15}\)
= \(\frac{2}{15}\)
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
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)
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???