Đáp án:
$1)
a)623\\
b) \dfrac{41}{12}\\
c)
\dfrac{-41}{12}\\
2)
a) x=6\\
b)x=\dfrac{-31}{6}\\
c)
{\left[\begin{aligned}x=\dfrac{1}{6}\\ x=\dfrac{-3}{2}\end{aligned}\right.}$
4)
a)
Oz nằm giữa hai tia Ox và Oy
b)
$\widehat{zOy}=40^{\circ}$
c)
Oz là tia phân giác của góc $\widehat{xOy}$
Giải thích các bước giải:
$1)
a)-127+323+427=(-127+427)+323=300+323=623\\
b) 10\dfrac{11}{12}-\dfrac{5}{7}.10\dfrac{1}{2}\\
=\dfrac{131}{12}-\dfrac{5}{7}.\dfrac{21}{2}\\
=\dfrac{131}{12}-\dfrac{15}{2}\\
=\dfrac{131}{12}-\dfrac{15.6}{12}\\
=\dfrac{41}{12}\\
c)
\dfrac{-3}{7}.\dfrac{-1}{12}+\dfrac{-3}{7}.\dfrac{11}{6}-\dfrac{8}{3}\\
=\dfrac{-3}{7}.\left ( \dfrac{-1}{12}+\dfrac{11}{6} \right )-\dfrac{8}{3}\\
=\dfrac{-3}{7}.\left ( \dfrac{-1}{12}+\dfrac{22}{12} \right )-\dfrac{8}{3}\\
=\dfrac{-3}{7}.\dfrac{21}{12}-\dfrac{8}{3}\\
=\dfrac{-3}{4}-\dfrac{8}{3}\\
=\dfrac{-9}{12}-\dfrac{32}{12}\\
=\dfrac{-41}{12}\\
2)
a)5x-13=17\\
\Leftrightarrow 5x=17+13=30\\
\Leftrightarrow x=6\\
b)\left ( x-\dfrac{5}{6} \right ):\dfrac{9}{2}=\dfrac{-4}{3}\\
\Leftrightarrow \left ( x-\dfrac{5}{6} \right ).\dfrac{2}{9}=\dfrac{-4}{3}\\
\Leftrightarrow x-\dfrac{5}{6} =\dfrac{-4}{3}.\dfrac{9}{2}\\
\Leftrightarrow x-\dfrac{5}{6} =-6\\
\Leftrightarrow x=-6+\dfrac{5}{6} \\
\Leftrightarrow x=\dfrac{-36}{6}+\dfrac{5}{6} \\
\Leftrightarrow x=\dfrac{-31}{6}\\
c)
\left | 3x+2 \right |=\dfrac{5}{2}\\
\Leftrightarrow {\left[\begin{aligned}3x+2=\dfrac{5}{2}\\ 3x+2=\dfrac{-5}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}3x=\dfrac{5}{2}-2\\ 3x=\dfrac{-5}{2}-2\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}3x=\dfrac{5}{2}-\dfrac{4}{2}\\ 3x=\dfrac{-5}{2}-\dfrac{4}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}3x=\dfrac{1}{2}\\ 3x=\dfrac{-9}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{6}\\ x=\dfrac{-3}{2}\end{aligned}\right.}$
4)
a)
Vì Oy và Oz cùng nằm trên 1 nữa mặt phẳng
mà $\widehat{xOy}<\widehat{xOz}$
nên $\widehat{xOz}+\widehat{zOy}=\widehat{xOy}$
Vậy Oz nằm giữa hai tia Ox và Oy
b)
$\widehat{zOy}=\widehat{xOy}-\widehat{xOz}=80^{\circ}-40^{\circ}=40^{\circ}$
c)
Vì $\widehat{xOz}=\widehat{zOy}=40^{\circ}$
Nên Oz là tia phân giác của góc $\widehat{xOy}$
