1 + 1.1
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a) \(1-\left(\frac{2x}{3}+2\right)=-1\cdot\frac{1}{3}\)
\(1-\frac{2}{3}x-2=-\frac{1}{3}\)
\(-\frac{2}{3}x-1=-\frac{1}{3}\)
\(-\frac{2}{3}x=\frac{2}{3}\)
\(x=-1\)
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b) \(\frac{2}{5}x-1\cdot\frac{1}{2}x+x=\frac{1}{3}\)
\(\left(\frac{2}{5}-\frac{1}{2}+1\right)x=\frac{1}{3}\)
\(\frac{9}{10}x=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{9}{10}\)
\(x=\frac{10}{27}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)..\left(1+\frac{1}{2019}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)
\(=\frac{2020}{2}=1010\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)
\(=\frac{2020}{2}\)
\(=1010\)