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\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5_{\cdot}+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(6-1\right)}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^2}{5}\)
\(=\dfrac{4}{5}\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
\(\frac{10.\left(4^6.9^5+6^9.120\right)}{8^4.3^{12}-6^{11}}\)
=\(\frac{2.5.\left[\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5\right]}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
=\(\frac{2^{13}.5.3^{10}+2^{13}.5^2.3^{10}}{2^{12}.3^{12}-3^{11}.2^{11}}\)
=\(\frac{2^{13}.5.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
=\(\frac{4.5.6}{3.5}\)
= 8
a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)
\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)
\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)
\(=9\dfrac{74}{99}-\dfrac{10}{9}\)
\(=\dfrac{965}{99}-\dfrac{10}{9}\)
\(=\dfrac{95}{11}\)
b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)
\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)
\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)
\(=\dfrac{1275}{22}\)
c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)
\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)
\(=\dfrac{89}{40}\)
d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)
\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)
\(=-\dfrac{25}{77}\)
e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
(câu này chờ mình một chút)
Câu e) anh rút thừa số chung là 2 cùng mũ số ra ngoài thì phân số thành số nguyên. Em nghĩ thế!
\(\frac{10.4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2.5.\left(2.2\right)^6.\left(3.3\right)^5+\left(2.3\right)^9.3.40}{\left(2.2.2\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2.5.2^6.2^6.3^5.3^5+2^9.3^9.3.40}{2^4.2^4.2^4.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{13}.5.3^{10}+2^{12}.5.3^{10}}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.5.3^{10}.\left(2+1\right)}{2^{11}.3^{11}\left(2.3-1\right)}=\frac{2^{12}.5.3^{11}}{2^{11}.5.3^{11}}=2\)
\(\dfrac{10.4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{5.2.2^{12}.3^{10}+3^9.2^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{5.2^{13}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\dfrac{2^{12}.3^{10}.5\left(2+1\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{11}.2.3^{10}.5.3}{2^{11}.3^{10}.3.5}=2\)
=(22)6.(32)5+(2.3)9.23.3.5/(23)4.312-(2.3)11
=212.310+29.39.23.3.5/212.312-211.311
=212.310+212.310.5/211.311(2.3-1)
=212.310(1+5)/211.311.5
=2.6/3.5
=4/5