a, \(4^7.3^4.9^6:6^{13}\)

\(=\left(2^{14}.3^4.3^{12}\right):\left(2^{13}.3^{13}\right)\)

\(=2^{14}:2^{13}.3^{16}:3^{13}\)

\(=2.3^3=54\)

b, \(2^3.3^2-5^{16}:25^7\)

\(=72-5^{16}:5^{14}\)

\(=72-5^2=47\)

4^7.3^4.9^6:6^13=4^7.3^4.(3^2)^6:6^13

=4^7.3^16:3^13.2^13

=(2^2)^7.3^16:3^13.2^13

=2^14.3^16:3^13.2^13

=2.3^3

=54

Mình làm câu đầu. Câu sau đang nghĩ

\(\frac{2.7.13}{26.35}=\frac{2.1.1}{2.5}=\frac{1}{5}\)

Ủng hộ tk Đúng nhé ! ^^ 

Ta có : \(C=\dfrac{4^6.3^4.9^5}{6^{12}}=\dfrac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}=\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=3^{4+10-12}=3^2=9\)

\(C=\dfrac{4^6\cdot3^4\cdot9^5}{6^{12}}=\dfrac{2^{12}\cdot3^4\cdot3^{10}}{3^{12}\cdot2^{12}}=3^2=9\)

\(C=\dfrac{4^6.3^4.9^5}{6^{12}}\) 

\(C=\dfrac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{\left(2.3\right)^{12}}\) 

\(C=\dfrac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}\) 

\(C=\dfrac{2^{12}.3^{14}}{2^{12}.3^{12}}\) 

\(C=3^2=9\)

\(\left(4096.81.729\right):2176782336\)

\(\left(331776.729\right):2176782336\)

\(\left(241864704\right):2176782336\)

\(\frac{4^6.3^4.9^3}{6^{12}}=\frac{2^{12}.3^4.3^6}{2^{12}.3^{12}}=\frac{2^{12}.3^{10}}{2^{12}.3^{12}}=\frac{1}{3^2}=\frac{1}{9}\)

=(2^12x3^10):2^12:3^12

=2^12:2^12x3^10:3^12

=3^(-2)

=1/9

B = 4^9.36+64^4
       16^4.100
B = 2^20.(9+16)
       2^18.5^2
B = 2^20.5^2
      2^18.5^2
B = 2^2
B = 4

D = 4^6.3^4.9^5
           6^12
D = 2^12.3^4.9^5
        2^12.3^12
D = 2^12.3^14
       2^12.3^12
D = 3^2
D = 9

\(\frac{4^9.36+64^4}{16^4.100}=\frac{4^{10}\left(9+8\right)}{4^{10}.25}=\frac{17}{25}\)

cảm ưn các bn nhìu

Bài 1:

\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)

\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)

\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)

Bài 2: 

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

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

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)