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A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
\(a,\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.7}=\frac{2.6}{3.7}=\frac{4}{7}\)
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
eo ôi t làm rồi mà bị xoá :v thôi t hướng dẫn :v
Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)
a/ \(\left(-0,125\right)^3:80^4=-\frac{1}{512}.80^4=-80000\)
b/ \(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=3\)
a, ( - 0, 125)^3 . 80^ 4
\(\left(-\frac{1}{8}\right)^3.80^4=\frac{-1^3}{8^3}\cdot8^4\cdot10^4\) = \(-8\cdot10^4=-8.1000=-8000\)
b,\(\frac{81^{11}.3^{17}}{27^{10}.9^{15}}=\frac{3^{4.11}.3^{17}}{3^{3.10}.3^{2.15}}=\frac{3^{44}.3^{17}}{3^{30}.3^{30}}=\frac{3^{44+17}}{3^{30+30}}=\frac{3^{61}}{3^{60}}=3\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)
