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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: \(B=\dfrac{2^{15}\cdot3^{30}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{3^3\cdot2\cdot2^{14}\cdot3^{14}\cdot3^{14}-2^2\cdot3\cdot2^{15}\cdot7^5}\)
\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)
\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\cdot\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)
\(A=\dfrac{115-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{30}}\)
\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
\(=\dfrac{5}{13}+3\)
\(=\dfrac{5}{13}+\dfrac{39}{13}\)
\(=\dfrac{44}{13}\)
\(=3\dfrac{5}{13}\)
CHÚC BN HC TỐT 
\(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=39/13+5/13=44/13
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)
\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)
\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)
\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)
\(A=\frac{5}{13}-3=-\frac{34}{13}\)
\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)
\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)
\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)
\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)
A= \(\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}\)+ \(\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{9}{10}}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
A= \(\dfrac{5}{13}\)+ \(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{3}{10}\right)}{1\left(\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{3}{10}\right)}\)
A= \(\dfrac{5}{13}+3\) = \(\dfrac{44}{13}\)
\(=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{2^{14}\cdot3^{14}\cdot3^3\cdot2\cdot3^{14}-2^{15}\cdot7^5\cdot2^2\cdot3}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-2^2\cdot7^5\right)}{2^{15}\cdot3^{31}-2^{17}\cdot7^5\cdot3}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-2^2\cdot7^5\right)}{2^{15}\cdot3\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)
Ta có: \(B=\dfrac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)
\(=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^{17}\cdot7^5}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}\)
\(=\dfrac{5}{3}\)
*Trả lời :
a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
=\(\dfrac{3}{4}.\left(-8\right)\)
= \(-6\)
b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
=\(10:\left(-\dfrac{5}{7}\right)\)
=\(-14\)
c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )
=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)
=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)
=\(\left(-1+1\right):\dfrac{7}{11}\)
\(=0:\dfrac{7}{11}\)
=0.
d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
a: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\left(\dfrac{1}{5}-\dfrac{3}{10}+\dfrac{1}{13}\right)}\)
=5/13+3
=5/13+39/13
=44/13
b: \(B=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{2\cdot3^3\cdot2^{14}\cdot3^{14}\cdot3^{14}-3\cdot2^2\cdot2^{15}\cdot7^5}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-7^5\cdot2^2\right)}{2^{15}\cdot3^{31}-3\cdot2^{17}\cdot7^5}\)
\(=\dfrac{2^{15}\cdot5\left(3^{30}-7^5\cdot2^2\right)}{2^{15}\cdot3\left(3^{30}-7^5\cdot2^2\right)}=\dfrac{5}{3}\)