\(\widehat{COB}+\widehat{DOB}=180^o.\\ \Rightarrow\widehat{COB}+4x=180^o.\\ \Rightarrow\widehat{COB}=180^o-4x.\) 

\(\widehat{AOC}+\widehat{DOA}=180^o.\\ \Rightarrow\widehat{AOC}+5x=180^o.\\ \Rightarrow\widehat{COB}=180^o-5x.\)

2 góc vuông không đối đỉnh:

\(\widehat{xAy};\widehat{x'Ay}.\)

\(\left|x-\dfrac{1}{2}\right|-\dfrac{1}{15}.\left(-3\right)^2=0,4.\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|-\dfrac{1}{15}.9=\dfrac{2}{5}.\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|-\dfrac{3}{5}=\dfrac{2}{5}.\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=1.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1.\\x-\dfrac{1}{2}=-1.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}.\\x=-\dfrac{1}{2}.\end{matrix}\right.\)

\(BT\Rightarrow\dfrac{x-3\sqrt{x}+x+6\sqrt{x}+9-x-9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-4}{\sqrt{x}-3}.\\ =\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-4}{\sqrt{x}-3}\\ =\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-4}{\sqrt{x}-3}.\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}+3}.\dfrac{\sqrt{x}-4}{\sqrt{x}-3}.\\ =\dfrac{\sqrt{x}-4}{\sqrt{x}+3}.\)

\(\dfrac{12}{-18}\) \(\in Q.\) Vì nó có dạng \(\dfrac{a}{b};a,b\in Z;b\ne0.\)

\(\dfrac{2}{3}-\dfrac{a}{b}+\dfrac{1}{4}=\dfrac{1}{6}.\\ \dfrac{11}{12}-\dfrac{a}{b}=\dfrac{1}{6}.\\ \dfrac{a}{b}=\dfrac{3}{4}.\)

\(\dfrac{2}{\sqrt{2}-1}+\sqrt{2}-\sqrt{18}=\dfrac{2}{\sqrt{2}-1}+\sqrt{2}-3\sqrt{2}=\dfrac{2}{\sqrt{2}-1}-2\sqrt{2}=\dfrac{2-2\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\dfrac{2-4+2\sqrt{2}}{\sqrt{2}-1}=\dfrac{-2+2\sqrt{2}}{\sqrt{2}-1}.\)

\(\widehat{zAn}=\widehat{xAy}=36^o\) (Đối đỉnh).

\(\widehat{zAy}=180^o\) (Góc bẹt).

\(\widehat{xAn}=180^o\) (Góc bẹt).

\(\widehat{xAz}+\text{​​}\widehat{xAy}=180^o.\\ \Rightarrow\widehat{xAz}+36^o=180^o.\\ \Rightarrow\widehat{xAz}=144^o.\)

\(\widehat{xAz}=\widehat{yAn}=144^o\) (Đối đỉnh).

\(\widehat{AOC}+\widehat{BOC}=180^o.\\ \Rightarrow45^o+\widehat{BOC}=180^o.\\\Rightarrow \widehat{BOC}=135^o.\)