tim gia tri nho nhat cua cac da thuc: a) P = x^2 - 2x + 5 b) Q = 2x^2 - 6x c) M = x^2 + y^2 - x + 6y + 10

Giải thích các bước giải:

a.P= (x-1)^2 +4 ≥ 4 min P=4 <=> x=1

b. 2.(x^2 -3x +9/4)-9/2 ≥ -9/2 . Min Q = -9/2 <=> x=3/2

c. (x-1/2)^2+ (y+3)^2 +3/4 ≥ 3/4 Min M = 3/4 <=> x=1/2 ; y= -3

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:

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