Question

khó kinh khủng luôn

Answer 1

Áp dụng t/c dtsbn:

\(\dfrac{11-5a}{2}=\dfrac{4b-28}{22}=\dfrac{4b-5a-17}{8a}=\dfrac{11-5a+4b-28-4b+5a+17}{2+22-8a}=\dfrac{0}{24+8a}=0\)

\(\Rightarrow\left\{{}\begin{matrix}11-5a=0\\4b-28=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{11}{5}\\b=7\end{matrix}\right.\)