2021^0-{15^2÷[275+(2^3.5^2-6.25)]}
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
160 - (23 . 52 - 6 . 25)
= 160 - (8 . 25 - 6 . 25
= 160 . [ 25 . (8-6) ]
= 160 . 50
= 8000
\(160-\left[2^3.5^2-6.25\right]\)
\(=160-\left[8.25-6.25\right]\)
\(=160-\left[25.\left(8-6\right)\right]\)
\(=160-\left[25.2\right]\)
\(=160-50\)
\(=110\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(160-\left(2^3.5^2-6.25\right)\)
\(=160-\left(8.25-6.25\right)\)
\(=160-\left[25.\left(8-6\right)\right]\)
\(=160-50\)
\(=110\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{2020}{2021}\)
\(\Leftrightarrow\frac{x+2}{2021}=1\)
\(\Leftrightarrow x=2019\)
Vậy \(x=2019\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{25}{4}+\dfrac{2}{15}\cdot\dfrac{3}{2}-1+\dfrac{1}{4}=\dfrac{26}{4}-1+\dfrac{1}{5}=\dfrac{22}{4}+\dfrac{1}{5}=\dfrac{114}{20}=\dfrac{57}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(160-\left(2^3.5^3-6.25\right)\)
\(=160-8.125+6.25\)
\(=160-1000+150\)
\(=210-1000\)
\(=-790\)
160-(2^3.5^3-6.25)
= 160 - ( 8.125.150 )
= 160 - 150000
= - 149840
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