\(Sosánh:\left(-32\right)^{27}và\left(-18\right)^{39}\)
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So Sánh : \(\left(-32\right)^{27}\)và\(\left(-18\right)^{39}\)
Ta có: \(32^{27}=\left(2^5\right)^{27}=2^{135}\)
\(16^{39}=\left(2^4\right)^{39}=2^{156}\)
mà \(2^{135}< 2^{156}\)
nên \(32^{27}< 16^{39}\)
mà \(16^{39}< 18^{39}\)
nên \(32^{27}< 18^{39}\)
\(\Leftrightarrow-32^{27}>-18^{39}\)
\(\Leftrightarrow\left(-32\right)^{27}>\left(-18\right)^{39}\)
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So sánh:
\(\left(-32\right)^{27}và\left(-18\right)^{39}\)
(-32)^27>(-18)^39 nha bn Cá là trong violympic lun
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Giúp tui với =)))
So sánh : \(\left(-32\right)^{27}va`\left(-18\right)^{39}\)
hồ trần nhi nói rứa thì coi như ko biết còn gì nữa
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Xem thêm câu trả lời
\(M=\left(-32\right)^{27}\) và \(N=\left(-18\right)^{41}\) so sánh M và N
so sanh
a) \(\left(\frac{1}{16}\right)^{200}\)va\(\left(\frac{1}{2}\right)^{1000}\)
b) (-32)27 va (-18)39
a, Có : (1/60)^200 = [(1/2)^4]^200 = (1/2)^800
Vì 0 < 1/2 < 1 nên (1/2)^800 > (1/2)^1000
=> (1/16)^200 > (1/2)^1000
Tk mk nha
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a) \(\left(\frac{1}{16}\right)^{200}=\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}\)
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a) \(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}\)
\(\left(\frac{1}{2}\right)^{1000}=\frac{1}{2^{1000}}\)
có : \(16^{200}=\left(2^4\right)^{200}=2^{800}\)
ta thấy \(2^{800}< 2^{1000}\)
\(\Rightarrow16^{200}< 2^{1000}\)
\(\Rightarrow\frac{1}{16^{200}}>\frac{1}{2^{1000}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{2}\right)^{100}\)
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Xem thêm câu trả lời
a) Aleft(frac{-5}{11}right).frac{7}{15}.left(frac{11}{-5}right).left(-30right)b) Bleft(-frac{1}{6}right).left(frac{-15}{19}right).left(frac{38}{45}right)c) C left(frac{-5}{9}right).frac{3}{11}+left(frac{-13}{18}right).frac{3}{11}d) D left(2frac{2}{15}.frac{9}{17}.frac{3}{32}right):left(frac{-3}{27}right)
Đọc tiếp
a) A=\(\left(\frac{-5}{11}\right).\frac{7}{15}.\left(\frac{11}{-5}\right).\left(-30\right)\)
b) B=\(\left(-\frac{1}{6}\right).\left(\frac{-15}{19}\right).\left(\frac{38}{45}\right)\)
c) C= \(\left(\frac{-5}{9}\right).\frac{3}{11}+\left(\frac{-13}{18}\right).\frac{3}{11}\)
d) D= \(\left(2\frac{2}{15}.\frac{9}{17}.\frac{3}{32}\right):\left(\frac{-3}{27}\right)\)
BT2: Tính nhanh
7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)
8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)
=\(\dfrac{-19}{34}.\dfrac{17}{19}+\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}-\dfrac{18}{7}.\dfrac{49}{18}\)
=\(\dfrac{1}{2}+\left(\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}\right)-7\)
=\(\dfrac{1}{2}+\left[\dfrac{49}{18}\left(\dfrac{-19}{34}+\dfrac{19}{34}\right)\right]-7\)
=\(\dfrac{1}{2}+0-7=\dfrac{-13}{2}\)
8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
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bài 1 : tính
a)\(\frac{-5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\) b) \(\left(\frac{3}{9}-\frac{9}{18}\right)+\frac{3}{6}-\left(\frac{1}{3}-\frac{1}{2}\right)-\frac{5}{15}\) c) \(\frac{9}{18}+\frac{16}{32}-\frac{12}{46}-\frac{9}{17}\) d) \(\left(\frac{14}{18}+\frac{-16}{27}\right)-\left(\frac{2}{3}-\frac{5}{15}\right)\)
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
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A = \(10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left(\sqrt{32}-\sqrt{8}-\sqrt{27}\right)\left(\sqrt{8}-\sqrt{32}-\sqrt{27}\right)\)
\(A=10-\left[-\sqrt{27}+\left(\sqrt{32}-\sqrt{8}\right)\right]\left[-\sqrt{27}-\left(\sqrt{32}-\sqrt{8}\right)\right]\)
\(A=10-\left[\left(-\sqrt{27}\right)^2-\left(\sqrt{32}-\sqrt{8}\right)^2\right]\)
\(A=10-\left(27-8\right)\)
\(A=-9\)
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