Đáp án:
1)
a) $
\widehat{yOm}=120^{\circ}$
2)
$a)\dfrac{2}{3}\\
b) \dfrac{-4}{7}\\
c)
\dfrac{21}{10}$
Giải thích các bước giải:
1)
a) $\widehat{mOn}=\widehat{xOy}-\widehat{xOm}-\widehat{nOy}\\
=180^{\circ}-60^{\circ}-30^{\circ}=90^{\circ}\\
\widehat{yOm}=\widehat{mOn}+\widehat{nOy}=90^{\circ}+30^{\circ}=120^{\circ}$
b)
Do $\widehat{mOn}=90^{\circ}$
Nên góc $\widehat{mOn}$ là góc vuông
c)
Do On là tia phân giác của góc $\widehat{yOt}$
$\Rightarrow \widehat{yOn}=\widehat{nOt}=30^{\circ}\\
\Rightarrow \widehat{yOt}=2.\widehat{yOn}=2.30^{\circ}=60^{\circ}\\
\widehat{xOt}=\widehat{xOy}-\widehat{yOt}=180^{\circ}-60^{\circ}=120^{\circ}\\
\Rightarrow \widehat{mOt}=\widehat{xOt}-\widehat{xOm}=120^{\circ}-60^{\circ}=60^{\circ}$
Do $\widehat{xOm}=\widehat{mOt}=60^{\circ}$
Nên Om là tia phân giác củ góc $\widehat{xOt}$
2)
$a)
\dfrac{2}{3}-\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{2}{3}\\
b) \dfrac{5}{7}-\dfrac{6}{5}.\dfrac{45}{42}=\dfrac{5}{7}-\dfrac{9}{7}=\dfrac{-4}{7}\\
c)
\dfrac{-12}{5}+1,5:\left ( 2-\dfrac{5}{3} \right )\\
=\dfrac{-12}{5}+\dfrac{3}{2}:\left (\dfrac{6}{3}-\dfrac{5}{3} \right )\\
=\dfrac{-12}{5}+\dfrac{3}{2}:\dfrac{1}{3}\\
=\dfrac{-12}{5}+\dfrac{3}{2}.3\\
=\dfrac{(-12).2}{10}+\dfrac{9.5}{10}\\
=\dfrac{21}{10}$
