x O y t A m C B

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 \(\widehat{xOt}=\widehat{tOy}\left(cmt\right)\)

\(\Rightarrow\widehat{ACO}=\widehat{BCO}\)

Ta có: \(\left\{{}\begin{matrix}\left|2y-27\right|^{2023}\ge0\forall y\\\left(3x+10\right)^{2024}\ge0\forall x\end{matrix}\right.\)

\(\Rightarrow\left|2y-27\right|^{2023}+\left(3x+10\right)^{2024}\ge0\forall x,y\)

Mà: \(\left|2y-27\right|^{2023}+\left(3x+10\right)^{2024}=0\)

nên: \(\left\{{}\begin{matrix}2y-27=0\\3x+10=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{27}{2}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: ...

\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)

\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)

\(\Rightarrow3x-12=-2x+4\)

\(\Rightarrow3x+2x=4+12\)

\(\Rightarrow5x=16\)

\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)

\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)