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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\right)+-\frac{1}{2}=\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\)\(-\frac{1}{2}\)
=\(\frac{6}{30}+\frac{10}{30}+\frac{9}{30}-\frac{15}{30}=\frac{6+10+9-15}{30}=\frac{10}{30}=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a) Ta có: \(\left|2x-5\right|\ge0\forall x\)
\(\Leftrightarrow-\left|2x-5\right|\le0\forall x\)
\(\Leftrightarrow-\left|2x-5\right|+3\le3\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Do \(\left|1+2x\right|\ge0\Rightarrow\dfrac{-1}{4}\left|1+2x\right|\le0\)
\(\Rightarrow A=2,25-\dfrac{1}{4}\left|1+2x\right|\le2,25\)
\(maxA=2,25\Leftrightarrow x=-\dfrac{1}{2}\)
b) Do \(\left|2x-3\right|\ge0\Rightarrow3+\dfrac{1}{2}\left|2x-3\right|\ge3\)
\(\Rightarrow B=\dfrac{1}{3+\dfrac{1}{2}\left|2x-3\right|}\le\dfrac{1}{3}\)
\(maxB=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-\frac{1}{3}+\frac{4}{5}\right|=\left|-3,2+\frac{2}{5}\right|\)
\(\Rightarrow x-\frac{1}{3}+\frac{4}{5}=-3,2+\frac{2}{5}\)
\(\Rightarrow x-\frac{1}{3}+\frac{4}{5}=-\frac{14}{5}\)
\(\Rightarrow x-\frac{1}{3}=-\frac{14}{5}-\frac{4}{5}\)
\(\Rightarrow x-\frac{1}{3}=-\frac{18}{5}\)
\(\Rightarrow x=\frac{-49}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\dfrac{1+5+5^2+5^3+5^4+...+5^{17}}{1+5^2+5^4+...+5^{16}}\)
Đặt tử số là B = 1 + 5 + 52 + 53 + 54 +...+ 517
5B = 5 + 52 + 53 + 54 +...+ 517 + 518
5B - B = 518 - 1
4B = 518 - 1
B = (518 - 1) : 4
Đặt mẫu số là C = 1 + 52 + 54 +...+ 516
52.C = 52 + 54 +...+ 516 + 518
25.C - C = 518 - 1
24C = 518 - 1
C = (518 - 1): 24
A = \(\dfrac{B}{C}\) = \(\dfrac{\dfrac{5^{18}-1}{4}}{\dfrac{5^{18}-1}{24}}\)
A = 6