Question 22: \(\text { Find } x \text { in an integer such that the expression } \mathrm{A}=\frac{5 x-19}{x-4} \text { sets the maximum value } \)

\(\begin{aligned} &\text { We have: } A=\frac{5 x-19}{x-4}=\frac{(5 x-20)+1}{x-4}=5 +\frac{1}{x-4} \text {. }\\ &\text { To } A_{\max } \text { then }(\mathrm{x}-4) \min \text {. } \\ &\text { TH1: } x-4<00 \Leftrightarrow x>4 \text {. To } \frac{1}{x-4} \max \text { then }(x-4)_{\min } \text { where } x \text { so } x-4=1\\ &\Leftrightarrow x=5\\ &\text { So } x=5 \text { then } A_{\max }=6 \text {. } \end{aligned}\ )