Đáp án:
$$\eqalign{
& a)\,\,{A_{\min }} = - 2 \Leftrightarrow x = {1 \over 4} \cr
& b)\,{B_{\min }} = - 5 \Leftrightarrow x = 0 \cr} $$
Giải thích các bước giải:
$$\eqalign{
& a)\,\,A = 4x - 2\sqrt x - 1 \cr
& A = {\left( {2\sqrt x } \right)^2} - 2\sqrt x + 1 - 2 \cr
& A = {\left( {2\sqrt x - 1} \right)^2} - 2 \cr
& \Rightarrow A \ge - 2 \Leftrightarrow {A_{\min }} = - 2 \cr
& \Leftrightarrow 2\sqrt x - 1 = 0 \Leftrightarrow \sqrt x = {1 \over 2} \Leftrightarrow x = {1 \over 4} \cr
& b)\,\,B = x + \sqrt x - 5 \cr
& B = x + 2.\sqrt x .{1 \over 2} + {1 \over 4} - {{21} \over 4} \cr
& B = {\left( {\sqrt x + {1 \over 2}} \right)^2} - {{21} \over 4} \cr
& \sqrt x + {1 \over 2} \ge {1 \over 2} \Leftrightarrow {\left( {\sqrt x + {1 \over 2}} \right)^2} \ge {1 \over 4} \cr
& \Leftrightarrow B \ge - 5 \cr
& \Rightarrow {B_{\min }} = - 5 \Leftrightarrow x = 0 \cr} $$
