Tinh (tinh nhanh neu co the)
\(2-\frac{2}{3+\frac{1}{2}}\) va \(\frac{1-\frac{1}{\frac{3}{4}+1}}{3}\)
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\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
= \(\frac{6}{10}-\frac{12}{30}\)
= \(\frac{18}{30}-\frac{12}{30}\)
= \(\frac{6}{30}=\frac{1}{5}\)
\(\frac{3}{4}.\frac{2}{5}-\frac{3}{10}.\frac{4}{3}\)
\(=\frac{3.2}{4.5}-\frac{3.4}{10.3}\)
\(=\frac{3}{10}-\frac{4}{10}=\frac{-1}{10}\)
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
\(\frac{x+5}{3}=\frac{5}{9}\)
=> ( x + 5 ) x 9 = 3 x 5
=> x + 5 = \(\frac{3\cdot5}{9}\)
=> x + 5 = \(\frac{5}{3}\)
=> x = \(\frac{-10}{3}\)
\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}\)
\(=\frac{16}{35}\)
\(2-\frac{2}{3+\frac{1}{2}}=2-\frac{2}{\frac{6}{2}+\frac{1}{2}}=2-\frac{2}{\frac{7}{2}}=2-2:\frac{7}{2}=2-\frac{4}{7}=\frac{10}{7}\)
\(\frac{1-\frac{1}{\frac{3}{4}+1}}{3}=1-\frac{1}{\frac{7}{\frac{4}{3}}}=1-\frac{1}{7}:\frac{4}{3}=1-\frac{3}{28}=\frac{25}{28}\)