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\(\frac{10^{15}.2^8}{2^{25}.25^7}=\frac{\left(2.5\right)^{15}.2^8}{2^{25}.\left(5^2\right)^7}=\frac{2^{15}.5^{15}.2^8}{2^{25}.5^{14}}=\frac{2^{23}.5^{15}}{2^{25}.5^{14}}=\frac{5}{2^2}=\frac{5}{4}\)
Dấu \(.\)là dấu nhân nha
\(\frac{10^{15}\times2^8}{2^{25}\times25^7}=\frac{2^{15}\times5^{15}\times2^8}{2^{25}\times5^{14}}=\frac{2^{23}\times5^{15}}{2^{25}\times5^{14}}\)
\(=\frac{5}{2^2}=\frac{5}{4}\)
#Hok_tốt ^^
\(E=\frac{4^9.9^5+6^9.2^6}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^9.\left(3^2\right)^5+6^9.64}{2^{10}.3^8+6^8.20}=\frac{2^{18}.3^{10}+6^9.64}{2^{10}.3^8+6^8.20}=\frac{2^8.3^2+6.2^4}{1.1+1.5}=\frac{2304+96}{6}=\frac{2400}{6}=400\)
a) Ta có: \(8^n:2^n=16^{2011}\)
\(\Leftrightarrow4^n=\left(4^2\right)^{2011}\)
\(\Leftrightarrow n=4022\)
b) Ta có: \(2^n+2^{n+3}=144\)
\(\Leftrightarrow2^n\left(1+2^3\right)=144\)
\(\Leftrightarrow2^n=16\)
hay n=4
\(8^n\div2^n=16^{2011}\)
\(\left(8\div2\right)^n=\left(4^2\right)^{2011}\)
\(4^n=4^{4022}\)
\(\Rightarrow n=4022\)
mình nghĩ ý b là
\(2^n+2^{n+3}=144\)
\(2^n+2^n\cdot2^3=144\)
\(2^n\left(1+8\right)=144\)
\(2^n\cdot9=144\)
\(2^n=16\)
\(2^n=2^4\)
\(\Rightarrow n=4\)
\(10^8\times2^8=\left(10\times2\right)^8=20^8\)
\(10^8:2^8=\left(10:2\right)^8=5^8\)
108 x 28 = (10x2)8 = 208
108 :28 = (10:2)8 = 58