Bài 1:
b) Ta có: \(\frac{\left[6-3\cdot\left(\frac{-1}{3}\right)^2+\sqrt{0.25}\right]}{\sqrt{0,\left(9\right)}}\)
\(=\frac{6-3\cdot\frac{1}{9}+0.5}{1}=\frac{6-\frac{1}{3}+\frac{1}{2}}{1}=\frac{36}{6}-\frac{2}{6}+\frac{3}{6}=\frac{37}{6}\)
Bài 2:
b) Ta có: \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\cdot\left(6x+1\right)\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\cdot6x+\frac{2}{3}\cdot1\)
\(\Leftrightarrow\frac{31}{60}-x=4x+\frac{2}{3}\)
\(\Leftrightarrow\frac{31}{60}-x-4x-\frac{2}{3}=0\)
\(\Leftrightarrow\frac{-3}{20}-5x=0\)
\(\Leftrightarrow5x=-\frac{3}{20}\)
\(\Leftrightarrow x=\frac{-3}{20}\cdot\frac{1}{5}=\frac{-3}{100}\)
Vậy: \(x=\frac{-3}{100}\)
https://i.imgur.com/ZmdvhR1.jpg
