\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

\(\frac{3}{5}-\frac{3}{25}-\frac{3}{125}\)

\(=\frac{75}{125}-\frac{15}{125}-\frac{3}{125}\)

\(=\frac{57}{125}\)

_______________________________

\(\frac{7}{5}-\frac{7}{25}-\frac{7}{125}\)

\(=\frac{175}{125}-\frac{35}{125}-\frac{7}{125}\)

\(=\frac{133}{125}\)

a) 1023579648

b) 9876541320

a) 1023457968

b) 9876543120

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B= \(1+3+3^2+...+3^{1000}\)

\(\Rightarrow3B=3+3^2+3^3+...+3^{1001}\)

\(\Rightarrow3B-B=\left(3+3^2+...+3^{1001}\right)-\left(1+3+...+3^{1000}\right)\)

\(\Rightarrow2B=3^{1001}-1\)

\(\Rightarrow B=\frac{3^{1001}-1}{2}\)

\(C=\)GHI LẠI ĐỀ NHA BN

\(2C=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(2C-C=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{...1}{2^{100}}\right)\)

\(C=2-\frac{1}{2^{100}}\)

K MÌNH NHA BN <3

bạn đang hỏi hay đang trả lời vậy???????????

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Giải
_  Ta có  \(\widehat{xOy}=\widehat{x'Oy'}=40^0\)( đối đỉnh) => \(\widehat{xOm}=\widehat{mOy}=\widehat{y'On}=\widehat{nOx'}=\frac{40^0}{2}=20^0\)
_  \(\widehat{x'Oy}=\widehat{xOy'}=180^0-40^0=140^0\)

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