\(\text{1, }\frac{27^4.9^3}{81^2}=\frac{\left(3^3\right)^4.\left(3^2\right)^3}{\left(3^4\right)^2}=\frac{3^{12}.3^6}{3^8}=3^{10}\)

\(\text{2, }\left(\frac{1}{5}\right)^{2002}.\left(-5\right)^{2000}=\frac{1}{5^{2002}}.5^{2000}=\frac{5^{2000}}{5^{2002}}=\frac{1}{5^2}=\frac{1}{5^2}\)

\(\text{3, }\frac{4^{11}.4^5}{2^{31}}=\frac{2^{22}.2^{10}}{2^{31}}=\frac{2^{32}}{2^{31}}=2\)

\(\text{4, }3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.81^2}{3^5.3^2}=\frac{3^2.3^8}{3^7}=\frac{3^{10}}{3^7}=3^3=27\)

\(\text{5, }4^2.\frac{25^2}{2^3.5^2}+32.125=\frac{2^4.5^4}{2^3.5^2}+2^5.5^3=2.5^2+2^5.5^2=5^2.\left(2+2^5.5\right)=25.\left(2+32.5\right)=25.162=4050\)

\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)

\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)

\(C=\frac{2}{3}\)

Bạn Cô nàng Thiên Bình làm đúng hết òi =.= 

a=7.[1/8+1/27-1/49]

   ------------------------

11.[1/8+1/27-1/49]

=7/11

cau b,c tuong tu nha h mk   

a)\(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)

\(A=\frac{7.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}{11.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}\).

\(A=\frac{7}{11}\)

b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)

\(B=\frac{\frac{8}{9}.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}{4.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}\)

\(B=\frac{8}{9}:4=\frac{2}{9}\)

c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)\(C=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}\)

C=\(\frac{2}{3}\)

=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)

=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)

=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)

=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)

=>x-3=0

=>x=3

A>5/3>5/4=>A>5/4 chứ mị

mk nhìn nhầm

a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)

\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(=\frac{1}{5}+\frac{2}{7}\)

\(=\frac{7}{35}+\frac{10}{35}\)

\(=\frac{17}{35}\)

Vậy \(A=\frac{17}{35}\)

b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)

\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)

\(=5.\frac{50}{671}\)

\(=\frac{250}{671}\)

Vậy \(B=\frac{250}{671}\)

a)

\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)

\(\Rightarrow A=\frac{17}{35}\)

b)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)

c)

\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow C=1-1-\frac{1}{25}\)

\(\Rightarrow C=\frac{1}{25}\)

Bài 1: 

\(A=\frac{24.47-22}{24+47.23}.\frac{5+\frac{5}{7}+\frac{5}{11}-\frac{5}{13}+\frac{5}{1001}}{6+\frac{6}{7}+\frac{6}{11}-\frac{6}{13}+\frac{6}{1001}}\)\(=\frac{47.23+47-22}{24.47.23}.\frac{5\left(1+\frac{1}{7}+\frac{1}{11.}-\frac{1}{13}+\frac{1}{1001}\right)}{6\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}+\frac{1}{1001}\right)}\) 

                                                                         \(=\frac{47.23+24}{24+47.23}.\frac{5}{6}\) 

                                                                         \(=1.\frac{5}{6}=\frac{5}{6}\) 

Bài 2: 

\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\) 

                                      \(=3^{28}-3^{27}-3^{26}\) 

                                      \(=3^{22}\left(3^6-3^5-3^4\right)\) 

                                      \(=3^{22}.405\) chia hết cho 405 

 =>đpcm