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a) \(\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{2^3}{3}=\frac{8}{3}\)
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\(\frac{9^3.4^6.8^2}{6^6.2^4}=\frac{\left(3^2\right)^3.\left(2^2\right)^6.\left(2^3\right)^2}{\left(3.2\right)^6.2^4}=\frac{3^6.2^{12}.2^6}{3^6.2^6.2^4}=\frac{2^{12}}{2^4}=2^8=256\)
\(\frac{9^3.4^6.8^2}{6^6.2^4}=\frac{\left(3^2\right)^3.\left(2^2\right)^6.\left(2^3\right)^2}{\left(2.3\right)^6.2^4}\) \(=\frac{3^6.2^{12}.2^6}{2^6.3^6.2^4}\) \(=\frac{2^{12}}{2^4}=2^8\)
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a, \(\frac{2^{15}.\left(-9\right)^4}{-6^3.8^3}=\frac{2^{15}.\left(-3.3\right)^4}{-\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^4.3^4}{-2^3.3^3.2^9}=\frac{2^{15}.3^8}{-2^{12}.3^3}=\frac{2^3.3^5}{-1}=-8.243=-1944\)
b, \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=\frac{1}{2^8}=\frac{1}{256}\)
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a,
\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)
b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)
c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
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\(E=\frac{4^9.9^5+6^9.2^6}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^9.\left(3^2\right)^5+6^9.64}{2^{10}.3^8+6^8.20}=\frac{2^{18}.3^{10}+6^9.64}{2^{10}.3^8+6^8.20}=\frac{2^8.3^2+6.2^4}{1.1+1.5}=\frac{2304+96}{6}=\frac{2400}{6}=400\)
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:V Làm sai hết rồi sai ngay từ bước đầu tiên.
\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)
\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{3}{20}\)
\(=\frac{-11}{12}\)
\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)
= \(-\frac{7}{30}\)
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a) \(2^5+8\left[\left(-2\right)^3:\frac{1}{2}\right]^0-\left(\frac{1}{2}\right)^3\times2+\left(-2\right)^3\)
\(=32+8\times1-\frac{1}{8}\times2+\left(-8\right)\)
\(=32+8-\frac{1}{4}+\left(-8\right)\)
\(=40-\frac{1}{4}+\left(-8\right)\)
\(=39\frac{3}{4}+\left(-8\right)\)
\(=31\frac{3}{4}\)
b vaf c mai minhf lamf, ht
=3/4
Tk mình với bạn ơi.
CHÚC BẠN HỌC TỐT ✓✓
\(\frac{9^5.2^{15}}{6^8.8^3}\)
\(=\frac{\left(3^2\right)^5.2^{15}}{\left(2.3\right)^8.\left(2^3\right)^3}\)
\(=\frac{3^{2.5}.2^{15}}{2^8.3^8.2^{3.3}}\)
\(=\frac{3^{10}.2^{15}}{2^8.2^9.3^8}\)
\(=\)\(\frac{3^{10}.2^{15}}{2^{17}.3^8}\)
\(=\frac{2^2}{3^2}=\frac{4}{9}\)