5: (-5/2)^2 + 2/15 . √ 9/4 - (-2021)^0 + | - 0,25 |
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5 x ( \(\dfrac{-5}{2}\) )2 + \(\dfrac{2}{15}\) x \(\sqrt{\dfrac{9}{4}}\) - ( -2022)0 + | -0,25|
= 5 x \(\dfrac{25}{4}\) + \(\dfrac{2}{15}\) x \(\dfrac{3}{2}\) - 1 + 0,25
= \(\dfrac{125}{4}\) + \(\dfrac{1}{5}\) - ( 1 - 0,25)
= 31,25 + 0,2 - 0,75
= 31,45 - 0,75
= 30,7
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ S = ( - 2/3 + 1) - (7/5 - 0,2) + )7/15 - 1 + 4/3) = - 2/3 + 1 - 7/5 + 1/5 +7/15 - 1 + 4/3 = 2/3 - 6/5 + 7/15 = 10/15 - 18/15 + 7/15 = - 1/15.
b/ S = 0,25 - 9- (- 3/4) - [ - 7/3 + (-9/2)] - 5/6 = 1/4 - 9 + 3/4 + 7/3 + 9/2 - 5/6 = - 8 + 14/6 + 27/6 - 5/6 = - 8 + 6 = - 2
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a)= 2021.2021-2020.(2021+1)
= 2021.(2020+1)-2020.(2021+1)
= (2021.2020)+2021-(2020.2021)-2020
= 1
b) B= (1+2-3-4)+(5+6-7-8)+(9+10-11-12)...........+(2017+2018-2019-2020)+2021
B= -4+(-4)+....................(-4)+2021
B= -4x505+2021
B= -2020 + 2021
B = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-7y=0\\11x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{11}\\y=\dfrac{x}{7}=\dfrac{5}{77}\end{matrix}\right.\)
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Lời giải:
a. Bạn cần viết đề bằng công thức toán để đề được rõ ràng hơn.
b. Ta có:
$(7y-x)^{2020}\geq 0$ với mọi $x,y$
$|5-11x|^{2021}\geq 0$ với mọi $x,y$
Do đó để tổng của chúng bằng $0$ thì:
$(7y-x)^{2020}=|5-11x|^{2021}=0$
$\Leftrightarrow x=\frac{5}{11}; y=\frac{5}{77}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{\dfrac{178}{495}}{\dfrac{623}{495}}-\dfrac{\dfrac{17}{60}}{\dfrac{119}{120}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\times\dfrac{2022}{2021}\)
\(M=0\times\dfrac{2022}{2021}\)
M=0
\(=\dfrac{25}{4}+\dfrac{2}{15}\cdot\dfrac{3}{2}-1+\dfrac{1}{4}=\dfrac{26}{4}-1+\dfrac{1}{5}=\dfrac{22}{4}+\dfrac{1}{5}=\dfrac{114}{20}=\dfrac{57}{10}\)