a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

a) \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{5}{21}\right):\dfrac{4}{5}+\left(\dfrac{5}{21}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0\)

b) \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\left(-\dfrac{3}{22}\right)+\dfrac{5}{9}:\left(-\dfrac{3}{5}\right)\)

\(=\dfrac{5}{9}:\left[\left(-\dfrac{3}{22}\right)+\left(-\dfrac{3}{5}\right)\right]\)

\(=\dfrac{5}{9}:\left(-\dfrac{81}{110}\right)\)

\(=-\dfrac{550}{729}\)

c) \(4^2.4^3:4^{10}\)

\(=\dfrac{4^5}{4^{10}}\)

\(=\dfrac{1}{4^5}\)

\(=\dfrac{1}{256}\)

d) \(\left(0,6\right)^5:\left(0,2\right)^6\)

\(=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^6}\)

\(=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^6}\)

\(=\dfrac{243}{0,2}\)

\(=1215\)

Mai mốt bạn đăng một lần ít thôi nha tại giờ khuya quá nên mình chỉ làm đến đây thôi =))

bạn viết cái này nó dễ hơn đó 

\(a\)) \(\left(4^2.4^3\right):2^{10}\)

   \(=4^5:2^{10}\)

   \(=\left(2^2\right)^5:2^{10}\)

   \(=2^{10}:2^{10}\)

   \(=1\)  

    \(b\)) \(\left(0,6\right)^5:\left(0,2\right)^6\)

    \(=\left(\frac{3}{5}\right)^5:\left(\frac{1}{5}\right)^6\)

     \(=\frac{3^5}{5^5}:\frac{1^6}{5^6}\)

      \(=\frac{3^5}{5^5}.5^6\)

        \(=\frac{3^5.5^6}{5^5}\)

        \(=3^5.5\)

         \(=243.5\)

          \(=1215\)

 \(c\)) \(\left(2^7.9^3\right):\left(6^5.8^2\right)\)

     \(=\left[2^7.\left(3^2\right)^3\right]:\left[\left(2.3\right)^5.\left(2^4\right)^2\right]\)

      \(=\left(2^7.3^6\right):\left[2^5.3^5.2^8\right]\)

       \(=\left(2^7.3^6\right).\left(\frac{1}{2^5.3^5.2^8}\right)\)

         \(=\frac{2^7.3^6.1}{\left(2^5.2^8\right).3^5}\)

           \(=\frac{2^7.3^6}{2^{13}.3^5}\)

           \(=\frac{3}{2^6}\)

           \(=\frac{3}{64}\)

Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha

1, 7A = 7+7^2+7^3+....+7^2008

6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1

=> A = (7^2008-1)/6

Tk mk nha

\(A=1+7+7^2+7^3+...+7^{2007}\)

\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)

\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)

\(\Rightarrow6A=7^{2008}-1\)

\(\Rightarrow A=\frac{7^{2008}-1}{6}\)

4b=4+4^2+4^3+...+4^101

4b-b=(4+4^2+...+4^101)-(1+4+4^2+...+4^100)

3b=4^101-1

b=(4^101-1):3

a) \(\frac{4^2.4^3}{2^{10}}\)\

\(=\frac{4^5}{\left(2^2\right)^{10}}=\frac{4^5}{4^{10}}=\frac{1}{4^5}\)

b) \(\frac{0,6^5}{0,2^6}=\frac{\left(0,2.3\right)^5}{0,2^6}=\frac{0,2^5.3^5}{0,2^6}=\frac{3^5}{0,2}=5.3^5\)

Là B bạn nha

Vì dãy số tự nhiên không giới hạn nha bạn

Câu B nha

TL: Câu B