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a, \(2^3.2^5=2^8=256\)
\(\left(-3\right)^9:\left(-3\right)^5=\left(-3\right)^4=81\)
\(\left(-6\right)^9.6^5=\left(-1\right)^9.6^9.6^5=\left(-1\right).6^{14}\\ \left(\dfrac{1}{2}\right)^5=\dfrac{1}{32}\)
b, \(\left(\dfrac{3}{5}\right)^6.\left(\dfrac{5}{3}\right)^6=\left(\dfrac{3}{5}. \dfrac{5}{3}\right)^6=1^6=1\\ \left(-\dfrac{7}{8}\right)^9:\left(\dfrac{7}{4}\right)^9=\left(-\dfrac{7}{8}:\dfrac{7}{4}\right)^9=\left(-\dfrac{1}{2}\right)^9=-\dfrac{1}{512}\\ \left(\left(-\dfrac{1}{2}\right)^2\right)^3=\left(\dfrac{1}{2}\right)^{...}\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)...\Rightarrow\left(\dfrac{1}{64}\right)=\left(\dfrac{1}{2}\right)^6\)
c, \(\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\left(\dfrac{2}{3}\right)^4\right)^2=\left(\left(\dfrac{2}{3}\right)^4\right)^{...}\Rightarrow\left(\dfrac{2}{3}\right)^8=\left(\left(\dfrac{2}{3}\right)^4\right)^2\\ \left(\dfrac{1}{3}\right)^{12}:\left(-\dfrac{3}{9}\right)^{12}=\left(\dfrac{1}{3}.\left(-3\right)\right)^{12}=\left(-1\right)^{12}=1\\ \left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{3}\right)^{10}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)
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Cho mình hỏi từ câu C trở xuống đc ko ạ tại mắt mình yếu nên nhìn không rõ ấy
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1) \(\left(\dfrac{-13}{17}-\dfrac{31}{52}\right)-\left(\dfrac{73}{52}-\dfrac{13}{17}+\dfrac{5}{6}\right)-\dfrac{3}{4}\)
\(=\dfrac{-13}{17}-\dfrac{31}{52}-\dfrac{73}{52}+\dfrac{13}{17}-\dfrac{5}{6}-\dfrac{3}{4}\)
\(=\left(\dfrac{-13}{17}+\dfrac{13}{17}\right)-\left(\dfrac{31}{52}+\dfrac{73}{52}\right)-\left(\dfrac{5}{6}+\dfrac{3}{4}\right)\)
\(=0-2-\dfrac{19}{12}\)
\(=-2-\dfrac{19}{12}\)
\(=\dfrac{-43}{12}\)
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a: \(=\dfrac{4}{7}+\dfrac{3}{7}\cdot\dfrac{-2}{3}\)
\(=\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{2}{7}\)
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a/
\(\widehat{xOt}=\widehat{tOy}=\dfrac{\widehat{xOy}}{2}=\dfrac{60^o}{2}=30^o\)
b/
\(\widehat{xAm}=\widehat{xOy}=60^o\)
Hai góc trên ở vị trí đồng vị => Am//Oy
c/
Ta có
Am//Oy (cmt) \(\Rightarrow\widehat{ACO}=\widehat{tOy}\) (góc so le trong)
BC//Ox (gt) \(\Rightarrow\widehat{BCO}=\widehat{xOt}\) (góc so le trong)
Mà \(\widehat{xOt}=\widehat{tOy}\left(cmt\right)\)
\(\Rightarrow\widehat{ACO}=\widehat{BCO}\)
=\(\frac{2.2.6}{2.2}+\frac{2.10}{2.13}+\frac{3.4}{13.4}+\frac{4.3}{13.4}\)\(+\frac{2.4}{4.2.9}+\frac{2.20}{9.2.7}\)
= \(6+\frac{10}{13}+\frac{3}{13}+\frac{1}{9}+\frac{20}{63}\)
= 6 +1+3/7
=52/7