Tính : A = 16^3 . 3^10 + 120 . 6^9 : 4^5 . 3^12 + 6^11
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\(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(A=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(A=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(A=\frac{2^{11}.3^{10}\left(2+2.5\right)}{2^{11}.3^{10}\left(2.3^2+3\right)}\)
\(A=\frac{2+10}{14}\)
\(A=\frac{6}{7}\)
\(16^3\cdot3^{10}+120\cdot6^9\)
\(=2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9\)
\(=2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5=2^{12}\cdot3^{10}\cdot6=2^{13}\cdot3^{11}\)
\(4^6\cdot3^{12}+6^{11}\)
\(=2^{12}\cdot3^{12}+2^{11}\cdot3^{11}\)
\(=2^{11}\cdot3^{11}\left(2\cdot3+1\right)=2^{11}\cdot3^{11}\cdot7\)
Ta có: \(\frac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)
\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\frac{2^2}{7}=\frac47\)
Ta có :
\(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(6+1\right)}=\frac{12}{21}=\frac{4}{7}\)
Chúc bạn học tốt ~
\(a.=5-3+12-4-16\)
\(=-2\)
\(b.=-6-\left(-12\right)+7-10\)
\(=6+7-10\)
\(=3\)