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\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}\)
\(=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)
\(\frac{2^7.9^3}{6^5.8^2}\)
\(=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
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\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(A=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
\(A=\frac{2}{3}\)
a,999/2610=111/290
b,374/506=17/23
c,3600-75/8400-175=3/7
d,9^14.25^5.8^7/18^12.625^3.24^3=3/25
a, Ta có:999/2610=111/290
b, Ta có:374/506=17/2
c,Ta có:3600-75/8400-175=48.75-75/48.175-175=75.(48-1)/175.(48-1)=75/175=3/7
d, Ta có:(3^2)^14.(5^2)^5.(2^3)^7/(2.3^2)^12.(5^4)^3.(2^3.3)^3
=3^28.5^10.2^21/2^12.3^24.5^12.2^9.3^3
=3^28.5^10.2^21/2^21.3^27.5^12
=3/5^2=3/25