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Dễ mà, bn nhân chia trước, cộng trừ sau, sau đó phá ngọăc rồi tính
\(=\frac{9^{15}.6^{30}}{27^{21}.8^{11}}\)
\(=\frac{\left(3^2\right)^{15}.3^{30}.2^{30}}{\left(3^3\right)^{21}.\left(2^3\right)^{11}}\)
\(=\frac{3^{30}.3^{30}.2^{30}}{3^{63}.2^{33}}\)
\(=\frac{3^{60}.2^{30}}{2^{63}.3^{33}}=\frac{1}{2^3.3^3}=\frac{1}{216}\)
làm mẫu một bài còn lại tương tự nha bn =)
\(\frac{9^{15}.6^{30}}{27^{21}.8^{11}}=\frac{\left(3^2\right)^{15}.\left(2.3\right)^{30}}{\left(3^3\right)^{21}.\left(2^3\right)^{11}}=\frac{3^{60}.2^{30}}{3^{63}.2^{33}}=\frac{1}{3^3.2^3}\)
\(\frac{45^{12}.49^7}{35^{13}.27^8}=\frac{\left(5.3^2\right)^{12}.\left(7^2\right)^7}{\left(5.7\right)^{13}.\left(3^3\right)^8}=\frac{5^{12}.3^{24}.7^{14}}{5^{13}.7^{13}.3^{24}}=\frac{7}{5}\)
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\(A=\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}=\frac{3^{16}.3^{10}.3^{15}:3^{12}}{3^{18}:3^6.3^{10}}=\frac{3^{29}}{3^{22}}=3^7\)
\(B=\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{2.55^2-9.55^2.55^{19}}{25^{10}}:\frac{5\left(21.7^{14}-19.7^{14}\right)}{7.7^{15}+3.7^{15}}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}:\frac{5.7^{14}.2}{7^{15}.10}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}.\frac{7^{15}.10}{5.7^{14}.2}\)Chịu ==
\(\frac{-12}{x}=\frac{4}{13}\)
\(\Rightarrow4x=13.\left(-12\right)=-156\)
\(\Rightarrow x=-156:4=-39\)
\(\frac{-15}{72}=\frac{125}{-3y}\)
\(\Rightarrow\frac{-15}{72}=\frac{-3000}{72y}\)
\(\Rightarrow-15y=-3000\)
\(\Rightarrow y=-3000:-15=200\)
\(\frac{49}{21}=\frac{-14z}{-7}\)
\(\Rightarrow\frac{49}{21}=\frac{42z}{21}\)
\(\Rightarrow49=42z\)
\(\Rightarrow z=\frac{49}{42}\)
\(\frac{21^{15}.7^{12}}{81^4.49^{13}}=\frac{\left(3.7\right)^{15}.7^{12}}{\left(3^4\right)^4.\left(7^2\right)^{13}}=\frac{3^{15}.7^{27}}{3^{16}.7^{26}}=\frac{7}{3}\)