a) \(T=\frac{9^{14}\times25^6\times8^7}{18^{12}\times625^3\times24^3}\)

       \(=\frac{\left(3^2\right)^{14}\times25^6\times\left(2^3\right)^7}{\left(2\times3^2\right)^{12}\times\left(25^2\right)^3\times\left(3\times2^3\right)^3}\)

       \(=\frac{3^{28}\times25^6\times2^{21}}{2^{12}\times3^{24}\times25^6\times3^3\times2^9}\)

       \(=\frac{3^{28}\times25^6\times2^{21}}{\left(2^{12}\times2^9\right)\times\left(3^{24}\times3^3\right)\times25^6}\)

       \(=\frac{3^{28}\times25^6\times2^{21}}{2^{21}\times3^{27}\times25^6}=3\)

b)  \(A=\frac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^9\times6^{19}-7\times2^{29}\times27^6}\)

        \(=\frac{5\times\left(2^2\right)^{15}\times\left(3^2\right)^9-2^2\times3^{20}\times\left(2^3\right)^9}{5\times2^9\times\left(2\times3\right)^{19}-7\times2^{29}\times\left(3^3\right)^6}\)

        \(=\frac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^9\times2^{19}\times3^{19}-7\times2^{29}\times3^{18}}\)

        \(=\frac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{28}\times3^{19}-7\times2^{29}\times3^{18}}\)

        \(=\frac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{28}\times3^{18}\times\left(5\times3-7\times2\right)}\)

        \(=\frac{2\times\left(10-9\right)}{15-14}=\frac{2\times1}{1}=2\)

bn nguyễn văn kiệt lm đug r

5 x 7 x 77 - 7 x 60 + 49 x 25 - 15 x 42

= (5 x 77) x 7 - 7 x 60 + 7 x 7 x 25 - 15 x 6 x 7

= 385 x 7 - 7 x 60 + 7 x (7 x 25) - 90 x 7

= 385 x 7 - 7 x 60 + 7 x 175 - 90 x 7

= 7 x (385 - 60 + 175 - 90)

= 7 x 410

= 2870

= 7 x 5 x 77 - 7 x 60 + 7 x 7 x 25 - 7 x 6 x 15

= 7 x 5 x 77 - 7 x 5 x 12 + 7 x 5 x 5 x 7 - 7 x 5 x 3 x 6

= 7 x 5 x ( 77 - 12 + 5 x 7 - 3 x 6 )

= 7 x 5 x ( 77 - 12 + 35 - 36 )

= 7 x 5 x ( 65 + 35 - 36 )

= 7 x 5 x ( 100 - 36 )

= 7 x 5 x 64

= 2240

1) 

a) \(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{24}.5^{12}.3^3.2^9}=\frac{3}{5^2}=\frac{3}{25}\)

Bài 2:

\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd};\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)

Vậy \(\frac{ab}{cd}=\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)

k nha 

mọi người giúp mình nha

a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)

b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)

c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)

d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)

\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)

\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)

a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=\frac{7^3.5^8}{7^2.5^8}=7\)

b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{5.3.5^4.3^8}=\frac{3^9.5^5}{5^5.3^9}=1\)

c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)

d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(-\frac{1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=1,22\)

Tham khảo tại đây:

https://olm.vn/hoi-dap/detail/80177168514.html

\(B=\frac{18\times123+9\times4567\times2+3\times5310\times6}{1+4+7+10+...+55+58-410}.\)

\(B=\frac{18\times123+9\times2\times4567+3\times6\times5310}{\left(1+4+7+10+....+55+58\right)-410}\)

\(B=\frac{18\times123+18\times4567+18\times5310}{\left(1+4+7+10+......+55+58\right)-410}\)

\(B=\frac{18\times\left(123+4567+5310\right)}{\left(1+4+7+10+....+55+58\right)-410}\)

\(B=\frac{18\times10000}{\left(1+4+7+10+....+55+58\right)-410}\)

Ta xét : 1 + 4 + 7 + 10 + .... + 55 + 58

Ta có : 4 - 1 = 3

            7 - 4 = 3

            10 - 4 = 3

           ................

            58 - 55 = 3

Vậy khoảng cách giữa 2 số liền nhau trong dãy số trên hơn kém nhau 3 đơn vị

Dãy số trên có tất cả số số hạng là :

( 58 - 1 ) : 3 + 1 = 20 ( số )

tổng của dãy số trên là :

( 58 + 1 ) x 20 : 2 = 590

Thay vào ta có :

\(B=\frac{18\times10000}{590-410}\)

\(B=\frac{180000}{180}\)

\(B=1000\)

Mình nhầm nhé : ( Mình quên chưa chia 2 )

1 + 4 + 7 + 10 + ... + 58 - 410

= ( 58 + 1 ) x [ ( 50 - 1 ) : 3 + 1 ]  : 2 - 410 

= 59 x 20 : 2 - 410

= 590 - 410 = 180

=> B = 18 000 : 180 = 100

\(A=\frac{0,8\times0,4\times1,25\times25+0,6524+0,3476}{10\times125\times4\times25\times8}\)

\(A=\frac{0,8\times1,25\times0,4\times25+0,6524+0,3476}{10\times4\times25\times8\times125}\)

\(A=\frac{1\times10+1}{10\times100\times1000}\)

\(A=\frac{11}{1000000}\)

\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{2^6}\)\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)

                                  \(=\left(\frac{3}{7}\right)^{21-12}\)\(=\left(\frac{3}{7}\right)^9\)

\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)

\(=\left(\frac{3}{7}\right)^9.\left(\frac{3}{7}\right)^{12}:\left[\left(\frac{3}{7}\right)^2\right]^6\)

\(=\left(\frac{3}{7}\right)^9.\left(\frac{3}{7}\right)^{12}:\left(\frac{3}{7}\right)^{12}\)

\(=\left(\frac{3}{7}\right)^9=\frac{3^9}{7^9}=\frac{19683}{40353607}\)

(3/7)²¹:(3/7²)⁶

(3/7)²¹:(3/7)¹²

(3/7)^21-12

(3/7)⁹