12^3 : (3^-4 * 64)
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\(\dfrac{\sqrt{4}}{3}\) - \(\dfrac{\sqrt{64}}{12}\)+ 3 \(\times\) \(\dfrac{\sqrt{1}}{27}\)
= \(\dfrac{2}{3}\) - \(\dfrac{8}{12}\) + 3 \(\dfrac{1}{27}\)
= \(\dfrac{2}{3}\) - \(\dfrac{2}{3}\) + \(\dfrac{1}{9}\)
= \(\dfrac{1}{9}\)
123:(3-4.64) = 33.43:(3-4.43) = 33.43:(\(\frac{1}{3^4}\).43)
=33.43:\(\frac{4^3}{3^4}\)= 33.43.34 :43= 37
=2187
\(12^3:\left(3^{-4}.64\right)\)
\(=1728:\left(\frac{1}{81}.64\right)\)
\(=1728:\frac{64}{81}\)
\(=2187\)
a: 2/3+3/4=11/12
9/4+3/5=45/20+12/20=57/20
2/5+4/7=14/35+20/35=34/34
3/5+4/3=9/15+20/15=29/15
b: 3/12+1/4=2/4=1/2
4/25+3/5=4/25+15/25=19/25
26/81+4/27=26/81+12/81=38/81
5/64+7/8=5/64+56/64=61/64
\(36+4\times y:3-12=64\)
\(36+4\times y\div3=64+12\)
\(36+4\times y\div3=76\)
\(4\times y\div3=76-36\)
\(4\times y\div3=40\)
\(4\times y=40\times3\)
\(4\times y=120\)
\(y=120:4\)
\(y=30\)
Ta có: \(64^{12}=\left(4^3\right)^{12}=4^{36}\)
\(S=4^0+4^1+...+4^{34}+4^{35}\)
\(\Rightarrow4S=4^1+4^2+...+4^{35}+4^{36}\)
\(\Rightarrow4S-S=4^{36}-4^0\)
\(\Rightarrow3S=4^{36}-1< 4^{36}\)
Vậy \(3S< 64^{12}\)