\(A=\left(4^2\right)^{17}.\frac{64^{36}}{8^{35}.32^{34}}\)
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\(A=\frac{\left(4^2\right)^{17}.64^{36}}{8^{35}.32^{34}}\)
\(A=\frac{\left(2^4\right)^{17}.\left(2^6\right)^{36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=\frac{2^{68}.2^{216}}{2^{105}.2^{170}}=\frac{2^{284}}{2^{275}}=2^9=512\)
\(A=64^{11}\cdot16^{13}=2^{66}\cdot2^{52}=2^{118}\)
\(B=32^{17}\cdot8^{19}=2^{85}\cdot2^{57}=2^{142}\)
Do đó: A<B
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)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
A = \(\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}\)
A = \(\left(-\frac{1}{4}\right)+\frac{5}{6}-\frac{7}{12}\)
A = \(\frac{7}{12}-\frac{7}{12}\)
A = \(0\).
Mình làm câu A thôi nhé.
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Câu a) bạn tham khảo tại đây nhé: Câu hỏi của Hằng Thanh.
Chúc bạn học tốt!
\(A=\left(4^2\right)^{17}.\frac{64^{36}}{8^{35}.32^{34}}\)
\(A=16^{17}.\frac{64^{2.36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=...6.\frac{...6^{36}}{2^{105}.2^{170}}\)
mình lười tính quá bạn. bạn làm nốt nha