\(=>\frac{2^{60}.2^{40}.3^7}{2^{50}.2^{30}.3^4}\)

\(=>\frac{2^{100}.3^7}{2^{80}.3^4}\)

\(=>\frac{2^{20}.3^3}{1}\)

\(\frac{8^2.4^5}{2^{20}}=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}\)

\(\frac{8^2.4^5}{2^{20}}=\frac{\left(2^3\right)^2.\left(2^2\right)^5}{2^{20}}=\frac{2^6.2^{10}}{2^{20}}=\frac{2^{16}}{2^{20}}=\frac{1}{2^4}=\frac{1}{16}\)

Cảm ơn bạn nhiều 

bạn có thể giải giúp mình một vài bào nữa k ?

\(=\frac{5^5\cdot\left(4.5\right)^3-5^4\cdot\left(4.5\right)^3+5^7\cdot4^5}{\left(5^3\right)^3\cdot4^5}=\frac{5^8.4^3-5^7.4^3+5^7.4^5}{5^9.4^5}=\frac{5^7.4^3.\left(5-1+4^2\right)}{5^7.4^3.\left(5^2.4^2\right)}\)

= \(\frac{4+4^2}{5^2.4^2}=\frac{4.5}{5^2.4^2}=\frac{1}{4.5}=\frac{1}{20}\)

Câu 1 : \(1,321338308x10^{-4}\)

Câu 2 : \(1316,572106\)

Câu 3 : \(1,641302619x10^{-13}\)

Ủng hộ nhé,tớ đang âm.

mấy câu này dễ

\(B=\frac{5}{2}+\left(\frac{4}{1.11}+\frac{3}{11.2}\right)+\left(\frac{1}{2.15}+\frac{13}{15.4}\right)\)

\(B=\frac{5}{2}+\frac{1}{11}.\left(4+\frac{3}{2}\right)+\frac{1}{15}\left(\frac{1}{2}+\frac{13}{4}\right)=\frac{5}{2}+\frac{1}{11}.\frac{11}{2}+\frac{1}{15}.\frac{15}{4}\)

=> \(B=\frac{5}{2}+\frac{1}{2}+\frac{1}{4}=\frac{10}{4}+\frac{2}{4}+\frac{1}{4}=\frac{13}{4}\)

Ta có

1/7.B = 5/2.7 + 4/7.11 + 3/11.14 + 1/14.15 + 13/15.28

1/7.B = 1/2 - 1/7 + 1/7 - 1/11 + 1/11 - 1/14 + 1/14 - 1/15 + 1/15 - 1/28

1/7.B = 1/2 - 1/28

1/7.B = 14/28 - 1/28

1/7.B = 13/28

B = 13/28 : 1/7

B = 13/28 . 7

B = 13/4

Ta có :

\(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(B=7\left(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\right)\)

   \(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\) 

     \(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)

      \(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)

      \(=7.\frac{13}{28}\)

      \(=\frac{13}{4}\)

Bài làm:

Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)

Cái còn lại tự CM

Ta có : 

\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\Rightarrow A< \frac{8}{9}\)(1)

Lại có \(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\Rightarrow A>\frac{2}{5}\)(2)

Từ (1) (2) => \(\frac{2}{5}< A< \frac{8}{9}\left(\text{ĐPCM}\right)\)

                         Bài làm :

Ta có :

\(A=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(A>\frac{1}{2}-\frac{1}{10}\)

\(A>\frac{2}{5}\left(1\right)\)

Ta cũng có  : 

\( A=\frac{1}{2.2}+\frac{1}{3.3}+......+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{8.9}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-......+\frac{1}{8}-\frac{1}{9}\)

\(A< 1-\frac{1}{9}\)

\(A< \frac{8}{9}\left(2\right)\)

\(\text{Từ (1) và (2) }\Rightarrow\frac{2}{5}< A< \frac{8}{9}\)

=> Điều phải chứng minh

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!