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\(\frac{10^3+2.5^3+5^3}{55}-\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)\(=\frac{2^3.5^3+2.5^3+5^3}{5.11}-\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{5^3\left(2^3+2+1\right)}{5.11}-\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{5^2.11}{11}-\frac{2.6}{3.5}=25-\frac{4}{5}=\frac{121}{5}\)
ta có : \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{4\left(4^5.9^5+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}=\dfrac{4\left(2^{10}.3^{10}+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}\)
\(=\dfrac{4\left(6^{10}+5.6^{10}\right)}{-6^{12}-6^{11}}=\dfrac{4.6^{11}}{-6^{11}\left(6+1\right)}=-\dfrac{4}{7}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=-\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}=\dfrac{-2}{3}\)
\(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)
\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}+\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{-6^{12}+6^{11}}\)
\(=\dfrac{2^{12}.3^{10}.6}{6^{11}-6^{12}}\)
\(=\dfrac{2^{12}.3^{10}.2.3}{6^{11}\left(1-6\right)}\)\(=\dfrac{2^{13}.3^{11}}{6^{11}.\left(-5\right)}\) \(=\dfrac{2^{11}.3^{11}.2^2}{6^{11}.\left(-5\right)}=\dfrac{6^{11}.4}{6^{11}.\left(-5\right)}=\dfrac{4}{-5}=-\dfrac{4}{5}\)
bài này mk làm ròi ak, tại do lỗi trang wed nên mới bị đăng lên
1. Thực hiện phép tính :
a. \(A=\left[6.\left(\dfrac{-1}{3}\right)-3.\left(\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(A=\left[-2-1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(A=-2:\left(-\dfrac{4}{3}\right)\)
Vậy : \(A=\dfrac{8}{3}\)
a, (5-5)-1 . \(\left(\dfrac{1}{2}\right)^{-2}\) . \(\dfrac{1}{10^5}\)
= 55 . \(\dfrac{1^{-2}}{2^{-2}}\) . \(\dfrac{1}{10^5}\)
= (55 . \(\dfrac{1}{10^5}\)) . \(\dfrac{1}{4}\)
= \(\dfrac{5^5}{10^5}\) . \(\dfrac{1}{4}\) = \(\left(\dfrac{1}{2}\right)^5\). \(\dfrac{1}{4}\)
= \(\dfrac{1}{32}.\text{}\dfrac{1}{4}\)= \(\dfrac{1}{128}\)
b: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{4}{5}\)
c: \(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}=10\cdot7=70\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(=\dfrac{5^3\left(2^3+2+1\right)}{55}-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=5^2-\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=25-\dfrac{2}{3}\cdot\dfrac{6}{5}\)
=25-4/5
=24,2