A = 163.310+120.69 phần 46.312+611

Tính A 

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NT
29 tháng 6 2021 lúc 21:28

Ta có: \(A=\dfrac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}+6^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\)

\(=\dfrac{2\cdot6}{3\cdot7}=\dfrac{12}{21}=\dfrac{4}{7}\)

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NL
29 tháng 6 2021 lúc 21:29

Ta có : \(A=\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.\left(1+6\right)}=\dfrac{2^2}{7}=\dfrac{4}{7}\)

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\(A=\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\) 

\(A=\dfrac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{11}.3+\left(2.3\right)^{11}}\) 

\(A=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{11}.3+2^{11}.3^{11}}\) 

\(A=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}.2.3+2^{11}.3^{11}}\) 

\(A=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}.6+2^{11}.3^{11}}\)  

\(A=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(6+1\right)}\)  

\(A=\dfrac{2.6}{3.7}\)

\(A=\dfrac{12}{21}=\dfrac{4}{7}\)

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