Bài 1:
a) \(\frac{8^5}{4^7}=\frac{\left(2^3\right)^5}{\left(2^2\right)^7}=\frac{2^{15}}{2^{14}}=2^1=2\)
b) \(\frac{49^2.7^8}{98.7^9}=\frac{\left(7^2\right)^2.7^8}{\left(2.7^2\right).7^9}=\frac{7^4}{2.7^2.7}=\frac{7^3}{2.7^2}=\frac{7}{2}\)
1
a)\(\frac{8^5}{4^7}=\frac{4^5.2^5}{4^7}=\frac{32}{16}=2\)
b)\(\frac{49^2.7^8}{98.7^9}=\frac{49^2.7^8}{2.49.7^9}=\frac{49}{2.7}=\frac{7}{2}\)
Bài 1) Rút gọn :
a) \(\frac{8^5}{4^7}=\frac{\left(2^3\right)^5}{\left(2^2\right)^7}=\frac{2^{3.5}}{2^{2.7}}=\frac{2^{15}}{2^{14}}=2\)
b) \(\frac{49^2.7^8}{98.7^9}=\frac{49.49.7^8}{49.2.7^8.7}=\frac{49}{2.7}=\frac{49}{14}=\frac{7}{2}=3,5\)
Bài 2 Chứng minh rằng :
a) \(11^6-11^5+11^4\)
\(=11^4.\left(11^2-11+1\right)\)
\(=11^4.\left(121-11-1\right)\)
\(=11^3.11.109⋮11\)
Vậy \(11^6-11^5+11^4⋮11\)(đpcm)
b) \(16^5+2^{19}-8^6\)
\(=\left(2^4\right)^5+2^{19}-\left(2^3\right)^6\)
\(=2^{4.5}+2^{19}-2^{3.6}\)
\(=2^{20}+2^{19}-2^{18}\)
\(=2^{18}.\left(2^2+2-1\right)\)
\(=2^{17}.2.5\)
\(=2^{17}.10⋮10\)
Vậy \(16^5+2^{19}-8^6⋮10\)(đpcm)
