Ta có : \(B_2+B_3=180^o\) ( 2 góc kề bù )

Mà : \(B_3=120^o\left(gt\right)\)

\(\Rightarrow B_2+120^o=180^o\Rightarrow B_2=60^o\)

\(\Rightarrow B_2=A_4\left(=60^o\right)\) ( 2 góc so le trong )

=> a // b ( t/c 2 đg thẳng // )

Vậy ...

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\left(k\ne0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\\z=7k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{k\left(2-5+7\right)}{k\left(2+10-7\right)}=\dfrac{4}{5}\)

Vậy A = \(\dfrac{4}{5}\)