Đáp án:
\(\eqalign{
& {A_{\max }} = 5 \Leftrightarrow x = 1 \cr
& {B_{\min }} = 2 \Leftrightarrow x \le - {2 \over 3} \cr
& {C_{\min }} = {5 \over 2} \Leftrightarrow x = 2019 \cr} \)
Giải thích các bước giải:
\(\eqalign{
& A = {{15} \over {2\left| {x - 1} \right| + 3}} \cr
& Ta\,\,co:\,\,\left| {x - 1} \right| \ge 0\,\,\forall x \cr
& \Rightarrow 2\left| {x - 1} \right| + 3 \ge 3\,\,\forall x \cr
& \Rightarrow A = {{15} \over {2\left| {x - 1} \right| + 3}} \le {{15} \over 3} = 5\,\,\forall x \cr
& {A_{\max }} = 5 \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1 \cr
& B = \left| {2 + 3x} \right| + 4 + 3x = \left[ \matrix{
2 + 3x + 4 + 3x = 6x + 6\,\,khi\,\,x \ge - {2 \over 3} \hfill \cr
- 2 - 3x + 4 + 3x = 2\,\,khi\,\,x < - {2 \over 3} \hfill \cr} \right. \cr
& Voi\,\,x \ge - {2 \over 3} \Leftrightarrow B = 6x + 6 \ge 2 \cr
& Vay\,\,{B_{\min }} = 2 \Leftrightarrow x \le - {2 \over 3} \cr
& C = {{\left| {x - 2019} \right| + 5} \over {2 + \left| {x - 2019} \right|}} \cr
& Dat\,\,t = \left| {x - 2019} \right| \Rightarrow t \ge 0 \cr
& C = {{t + 5} \over {2 + t}} = 1 + {3 \over {2 + t}} \cr
& t \ge 0 \Leftrightarrow 2 + t \ge 2 \Leftrightarrow {3 \over {2 + t}} \le {3 \over 2} \cr
& \Rightarrow C \le {5 \over 2} \Rightarrow {C_{\min }} = {5 \over 2} \Leftrightarrow t = 0 \Leftrightarrow x = 2019 \cr} \)
