(-27)^5:32^3
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2/3 x 4/5 + 4/5 x 8/3 = 4/5 x (2/3 + 8/3) = 4/5 x 10/3 = 8/3
27/32 x 16/9 -27/32 x7/9 + 27/32
= 27/32 x (16/9 - 7/9 + 1 )
=27/32 x 2
=27/16
\(\frac{2}{3}\) x \(\frac{4}{5}\) + \(\frac{4}{5}\) x \(\frac{8}{3}\)
=\(\frac{4}{5}\) x ( \(\frac{2}{3}\) + \(\frac{8}{3}\) )
= \(\frac{4}{5}\) x \(\frac{10}{3}\)
= \(\frac{40}{15}\) = \(\frac{8}{5}\)
a) Ta có \(0,625^{200}=\left(\dfrac{5}{8}\right)^{200}\) và \(0,5^{1000}=\left(\dfrac{1}{2}\right)^{1000}=\left(\dfrac{1}{2}\right)^{5.200}\) \(=\left[\left(\dfrac{1}{2}\right)^5\right]^{200}\) \(=\left(\dfrac{1}{32}\right)^{200}\). Mà hiển nhiên \(\left(\dfrac{5}{8}\right)^{200}>\left(\dfrac{1}{32}\right)^{200}\) nên suy ra \(0,625^{200}>0,5^{1000}\)
b) Ta thấy \(\left(-32\right)^{27}< 0\) trong khi \(\left(-27\right)^{32}>0\) nên đương nhiên \(\left(-32\right)^{27}< \left(-27\right)^{32}\)
c) Ta thấy \(-\dfrac{3}{2}>-2\) nên \(\left(-\dfrac{3}{2}\right)^5>\left(-2\right)^5\)
\(=\left(\dfrac{-12+5}{32}\right):\dfrac{-4}{5}+\dfrac{27}{125}+\dfrac{-5}{8}+\dfrac{27}{32}\cdot\dfrac{5}{4}\)
\(=\dfrac{-7}{32}\cdot\dfrac{-5}{4}+\dfrac{27}{125}+\dfrac{-5}{8}+\dfrac{135}{128}\)
\(=\dfrac{100}{128}+\dfrac{27}{125}-\dfrac{5}{8}\)
\(=\dfrac{1489}{4000}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
32/5 : 32/25 - 27/37 * 32/3
32/5 * 25/32 - 27/37 * 32/3
5 - 288/37
-103/37
\(\left(-27\right)^5:32^3\)
\(=\left[\left(-3\right)^3\right]^5:\left(2^5\right)^3\)
\(=\left(-3\right)^{15}:2^{15}\)
\(=\left(-3:2\right)^{15}\)
\(=\left(-\dfrac{3}{2}\right)^{15}\)
\(\left(-27\right)^5\):\(32^3\)
=-14348907:32768
=-437,8938904