\(\dfrac{8}{27}\)=\(\dfrac{2^n}{3\times3^m}\)
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a) \(\dfrac{2}{5}=\dfrac{2\times3}{5\times3}=\dfrac{6}{15}=\dfrac{2}{5}\)
\(\dfrac{4}{7}=\dfrac{4\times2}{7\times2}=\dfrac{8}{14}=\dfrac{4}{7}\)
\(\dfrac{13}{54}=\dfrac{13\times3}{54\times3}=\dfrac{39}{162}=\dfrac{13}{54}\)
b) \(\dfrac{8}{20}=\dfrac{8:4}{20:4}=\dfrac{2}{5}\)
\(\dfrac{10}{16}=\dfrac{10:2}{16:2}=\dfrac{5}{8}\)
\(\dfrac{25}{65}=\dfrac{25:5}{65:5}=\dfrac{5}{13}\)
\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)
a) \(\dfrac{9}{11}\times8=\dfrac{9\times8}{11}=\dfrac{72}{11}\)
b) \(\dfrac{4}{5}\times1=\dfrac{4\times1}{5}=\dfrac{4}{5}\)
c) \(\dfrac{15}{8}\times0=\dfrac{15\times0}{8}=\dfrac{0}{8}=0\)
a: 9/11*8=(9*8)/11=72/11
b: 4/5*1=(4*1)/5=4/5
c: 15/8*0=(15*0)/8=0/8=0
\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)
\(=\dfrac{1}{2}-\dfrac{1}{2450}\)
\(=\dfrac{612}{1225}\)
\(\text{#}Toru\)
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
n=3
m=2
\(\dfrac{8}{27}=\dfrac{2^n}{3\text{x}3^m}\)
=>\(\dfrac{2^3}{3^3}=\dfrac{2^n}{3\text{x}3^m}\)
n=3và \(3\text{x}3^2=3^{1+2}\)=>m=2
Vậyn=3,m=2