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Ta có:
B = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)
C = \(\frac{2^{10}.39+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(39+65\right)}{2^8.104}=\frac{2^{10}.104}{2^8.104}=4\)
Ta thấy : 3 < 4 => B < C
So Sánh: B = \(\frac{^{3^{10}.11+3^{10}.5}}{3^9.2^4}\) và C= \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
Ta có:
B=\(\frac{3^{10}.\left(11+5\right)}{3^9.2^4}\) = \(\frac{3^{10}.16}{3^9.2^4}\)= \(\frac{3^9.3.16}{3^9.16}\)= 3
C=\(\frac{2^{10}.\left(13+65\right)}{2^8.104}\) =\(\frac{2^{10}.78}{2^8.104}\) = \(\frac{2^8.2^2.78}{2^8.104}\)= \(\frac{4.78}{104}\) = \(\frac{4.78}{4.26}\)=\(\frac{78}{26}\)=3
=>B=C
a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)
b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)
c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)