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a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)
b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)
=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)
= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)
c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)
= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)
= 2 + 1=3
d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)
= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(\Rightarrow A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(\Rightarrow A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(\Rightarrow A=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-4\right)}{5^9.7^8\left(1+2^3\right)}\)
\(\Rightarrow A=\frac{2}{3.4}-\frac{5.\left(-3\right)}{9}\)
\(\Rightarrow A=\frac{1}{3}-\frac{-15}{9}\)
\(\Rightarrow A=\frac{1}{3}+\frac{5}{3}\)
\(\Rightarrow A=\frac{6}{3}=2\)
Vậy \(A=2\)
a) A = \(\frac{45^{10}.5^{10}}{75^{10}}=\frac{\left(5.3^2\right)^{10}.5^{10}}{\left(5^2.3\right)^{10}}=\frac{5^{10}.3^{20}.5^{10}}{5^{20}.3^{10}}=\frac{5^{20}.3^{10}}{5^{20}}=3^{10}=59049\)
b) B = \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2.2^2\right)^5}{\left(0,2.2\right)^6}=\frac{\left(0,2\right)^5.2^{10}}{\left(0,2\right)^6.2^6}=\frac{2^4}{0,2}=\frac{16}{0,2}=80\)
c) C = \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
\(A=\frac{\left(9^{10}×5^{10}×5^{10}\right)}{\left(25^{10}×3^{10}\right)}\)
\(A=\frac{\left(3^{20}×5^{20}\right)}{\left(5^{20}×3^{10}\right)}\)
\(A=\frac{3^{20}}{3^{10}}\)
\(A=3^{10}\)
\(\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{10}{7}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}=\frac{\left(\frac{4293}{324}-\frac{708}{324}-\frac{3510}{324}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{30}{21}+\frac{70}{21}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}=\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:\left(-\frac{41}{21}\right)}\)=\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}=100:\left(-\frac{100}{4}\right)=-4\)
\(30+\frac{14}{5}:\left(\frac{24}{150}-\frac{270}{150}-\frac{25}{150}\right)=30+\frac{14}{5}:\left(-\frac{271}{150}\right)=30+\left(-\frac{420}{271}\right)=\frac{7710}{271}\)
\(C=\left[-45^{10}.\left(-5\right)^{10}\right]:75^{10}\)
\(C=\left(-45^{10}.5^{10}\right):75^{10}\)
\(C=\left(-45.5\right)^{10}:75^{10}\)
\(C=\left(-225\right)^{10}:75^{10}\)
\(C=\left(-225:75\right)^{10}=\left(-3\right)^{10}=3^{10}\)
\(B=\left(7^4-7^3\right)^3:343^{12}\)
\(B=\left[7^3.\left(7-1\right)\right]^3:\left(7^3\right)^{12}\)
\(B=\left(7^3.6\right)^3:7^{36}\)
\(B=7^9.6^3:7^{36}\)
\(B=6^3:7^{27}\)