\(\orbr{-20.83\times0.2}+\left(-9.17\right).0.2]\div[2.47\times0.5\times\left(-3.53\right)\times0.5]\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[3,5+10.\left(0,4\right)^2\right]:\left[\left(0,5\right)^2-\left(\frac{1}{5}\right)^3+2,758\right]\)
\(=\left(3,5+1,6\right):\left[\frac{1}{4}-\frac{1}{125}+2,758\right]\)
\(=5,1:\left[\frac{125}{500}-\frac{4}{500}+\frac{1379}{500}\right]\)
\(=5,1:3\)
\(=1,7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\left(-20,83\right).0,2+\left(-9,17\right).0,2\right]:\left[2,47.0,5-\left(-3.53\right).0,5\right]\)
\(=\left\{0,2.\left[\left(-20,83\right)+\left(-9,17\right)\right]\right\}:\left[2,47.0,5+3,53.0,5\right]\)
\(=\left[0,2.\left(-30\right)\right]:\left[0,5.\left(2,47+3,53\right)\right]\)
\(=\left(-6\right):\left(0,5.6\right)\)
\(=\left(-6\right):3\)
\(=\left(-2\right)\)
[ (-20,83) . 0,2 + (-9,17) . 0,2 ] : [ 2,47 . 0,5 - (-3.53) . 0,5 ]
= [\(\frac{-2083}{500}\) + \(\frac{-917}{500}\) ] [ \(\frac{247}{200}-\frac{-353}{200}\) ]
= -6 : 3 = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\left(\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)
\(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{5^6}{7^6}:\dfrac{5^8}{7^6}=\dfrac{1}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,75-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=\frac{2}{3}\)
\(\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=4\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}=\frac{6}{5}\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]=1\)
\(\left(3,72-0,02.x\right)=\frac{37}{10}\)
\(0,02.x=0,02\)
\(x=1\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(\Rightarrow\frac{8}{3}:\left\{\left[\left(\frac{93}{25}-\frac{1}{50}.x\right)\frac{10}{37}\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{1}{5}\)
\(\Rightarrow\left\{\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{8}{3}:\frac{1}{5}=\frac{40}{3}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}=\frac{40}{3}+\frac{7}{15}=\frac{69}{5}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}=\frac{69}{5}-\frac{14}{5}=11\)
\(\Rightarrow\frac{93}{25}-\frac{1}{50}.x=11.\frac{5}{6}=\frac{55}{6}\)
\(\Rightarrow\frac{1}{50}.x=\frac{93}{25}-\frac{55}{6}=\frac{-817}{150}\)
\(\Rightarrow x=\frac{-817}{150}:\frac{1}{50}=\frac{-817}{3}\)
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