Ta có: \(B=\dfrac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)

\(=\dfrac{2^{15}\cdot5\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}\)

\(=\dfrac{5}{3}\)

Mik ko chắc lắm đâu nha(/Mik làm tắc nên mất phấn tắc ra TSNT)

= \(\dfrac{5}{2}\)

bạn THAM KHẢO thôi nhé!

Kq = 5/2 nha bạn câu này đã có người hỏi rồi bạn tự tham khảo nhé 

\(B=\frac{5.2^{15}.3^{30}-5.7^5.2^5.2^{12}}{5^4.2^{14}.3^{14}.3^7.3^7-2^2.3.2^{15}.7^5}\)

=> \(B=\frac{5.2^{15}.3^{30}-5.7^5.2^{17}}{5^4.2^{14}.3^{28}-2^{17}.3.7}\)

=> \(B=\frac{5.2.3^{30}-5.7^5.2^3}{5^4.3^{28}-2^3.3.7}=0,144\)

\(=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{14}\cdot3^{14}\cdot3^3\cdot2\cdot3^{14}-2^{15}\cdot7^5\cdot2^2\cdot3}\)

\(=\dfrac{2^{15}\cdot5\left(3^{30}-2^2\cdot7^5\right)}{2^{15}\cdot3^{31}-2^{17}\cdot7^5\cdot3}\)

\(=\dfrac{2^{15}\cdot5\left(3^{30}-2^2\cdot7^5\right)}{2^{15}\cdot3\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)

đề sai

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)

\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)

\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)

\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)

\(A=\frac{5}{13}-3=-\frac{34}{13}\)

\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)

\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)

\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)

1: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)

=5/13+3

=5/13+39/13

=44/13

2: \(B=\dfrac{2^{15}\cdot3^{30}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{3^3\cdot2\cdot2^{14}\cdot3^{14}\cdot3^{14}-2^2\cdot3\cdot2^{15}\cdot7^5}\)

\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)

\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\cdot\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)