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SS
26 tháng 6 2019
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
26 tháng 2 2020
a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)
⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)
hay x=5
Vậy: x=5
b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)
⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)
⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)
hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)
⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)
Vậy: \(x=\frac{29}{10}\)
c) \(10\cdot3^{x+2}-3^x=89\)
\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)
\(\Leftrightarrow3^x\left(90-1\right)=89\)
\(\Leftrightarrow3^x=1\)
hay x=0
Vậy: x=0
d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)
⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)
\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)
\(\Leftrightarrow2x-\frac{19}{27}=0\)
\(\Leftrightarrow2x=\frac{19}{27}\)
hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)
Vậy: \(x=\frac{19}{54}\)
e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)
\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
6 tháng 8 2020
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
5 tháng 12 2018
\(\text{a,}\frac{2}{13}.\frac{-5}{3}+\frac{11}{13}.\frac{-5}{3}=-\frac{5}{3}\left(\frac{2}{13}+\frac{11}{13}\right)\)
\(=\frac{-5}{3}.\frac{13}{13}\)
\(=-\frac{5}{3}\)
\(\text{b,}\left(-\frac{1}{3}\right)^2+\left(-\frac{1}{3}\right)^3.27+\left(\frac{-2017}{2018}\right)^0=\frac{1}{9}-\frac{1}{27}.27+1\)
\(=\frac{1}{9}-1+1\)
\(=\frac{1}{9}\)
\(\text{c,}1,2-\sqrt{\frac{1}{4}}:1\frac{1}{20}+\left|\frac{3}{4}-1,25\right|-\left(\frac{-3}{2}\right)^2=\frac{6}{5}-\frac{1}{2}:\frac{21}{20}+\left|\frac{3}{4}-\frac{5}{4}\right|-\frac{9}{4}\)
\(=\frac{6}{5}-\frac{10}{21}+\frac{1}{2}-\frac{9}{4}\)
\(=\frac{-431}{420}\)
a) \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)
b) \(\frac{3^2}{0,375^2}=\left(\frac{3}{0,375}\right)^2=8^2=64\)
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