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\(A=\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+...+\dfrac{10}{2^{10}}\)
\(2A=\dfrac{1}{1}+\dfrac{2}{2}+\dfrac{3}{4}+...+\dfrac{10}{2^9}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{4}+...+\dfrac{10}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{2}{4}+...+\dfrac{10}{2^{10}}\right)\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^9}-\dfrac{10}{2^{10}}\)
\(B=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^9}\)
\(2B=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\)
\(2B-B=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^9}\right)\)
\(B=2-\dfrac{1}{2^9}\)
Suy ra \(A=B-\dfrac{10}{2^{10}}=2-\dfrac{1}{2^9}-\dfrac{10}{2^{10}}=\dfrac{509}{256}\)
a) \(\dfrac{-7}{5}\) . \(\dfrac{4}{23}\) + \(\dfrac{4}{23}\) . \(\dfrac{2}{5}\)
= \(\dfrac{4}{23}\) . ( \(\dfrac{-7}{5}\) + \(\dfrac{2}{5}\) )
= \(\dfrac{4}{23}\) . -1
= \(\dfrac{-4}{23}\)
b) \(\dfrac{-6}{7}\) . \(\dfrac{3}{5}\) +\(\dfrac{-6}{7}\) .\(\dfrac{2}{5}\) - \(1\dfrac{1}{3}\)
= \(\dfrac{-6}{7}\) . \(\dfrac{3}{5}\) +\(\dfrac{-6}{7}\) .\(\dfrac{2}{5}\) - \(\dfrac{4}{3}\)
= \(\dfrac{-6}{7}\) . (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\) ) - \(\dfrac{4}{3}\)
= \(\dfrac{-6}{7}\) . 1 - \(\dfrac{4}{3}\)
= \(\dfrac{-6}{7}\) - \(\dfrac{4}{3}\)
= \(\dfrac{-18}{21}\) - \(\dfrac{28}{21}\)
= \(\dfrac{-18}{21}\) + \(\dfrac{-28}{21}\)
= \(\dfrac{-46}{21}\)
- Tất cả các đoạn thẳng có trong hình vẽ là : MP ; MQ ; MN ; NQ ; PQ ; NP
\(\Rightarrow\) Chọn \(B\)
_HT_