Đáp án:
\(\eqalign{
& a)\,{P_{\min }} = 4 \Leftrightarrow x = 1 \cr
& b)\,\,{Q_{\min }} = - {9 \over 2} \Leftrightarrow x = {3 \over 2} \cr
& c)\,\,{M_{\min }} = {3 \over 4} \Leftrightarrow \left\{ \matrix{
x = {1 \over 2} \hfill \cr
y = - 3 \hfill \cr} \right. \cr} \)
Giải thích các bước giải:
$$\eqalign{
& a)\,\,P = {x^2} - 2x + 5 \cr
& P = {x^2} - 2x + 1 + 4 \cr
& P = {\left( {x - 1} \right)^2} + 4 \cr
& P \ge 4 \Rightarrow {P_{\min }} = 4 \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x - 1 = 0 \Leftrightarrow x = 1 \cr
& b)\,\,Q = 2{x^2} - 6x \cr
& Q = 2\left( {{x^2} - 3x} \right) \cr
& Q = 2\left( {{x^2} - 2.x.{3 \over 2} + {9 \over 4}} \right) - {9 \over 2} \cr
& Q = 2{\left( {x - {3 \over 2}} \right)^2} - {9 \over 2} \cr
& \Rightarrow Q \ge - {9 \over 2} \Rightarrow {Q_{\min }} = - {9 \over 2} \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x = {3 \over 2} \cr
& c)\,\,M = {x^2} + {y^2} - x + 6y + 10 \cr
& M = \left( {{x^2} - x + {1 \over 4}} \right) + \left( {{y^2} + 6y + 9} \right) + {3 \over 4} \cr
& M = {\left( {x - {1 \over 2}} \right)^2} + {\left( {y + 3} \right)^2} + {3 \over 4} \cr
& \Rightarrow M \ge {3 \over 4} \Rightarrow {M_{\min }} = {3 \over 4} \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \left\{ \matrix{
x = {1 \over 2} \hfill \cr
y = - 3 \hfill \cr} \right. \cr} $$
