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 27/09/2021 Bởi Hailey 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
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: $$\eqalign{ & a)\,\,P = {x^2} – 2x + 5 \cr & P = {x^2} – 2x + 1 + 4 \cr & P = {\left( {x – 1} \right)^2} + 4 \cr & P \ge 4 \Rightarrow {P_{\min }} = 4 \cr & Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1 \cr & b)\,\,Q = 2{x^2} – 6x \cr & Q = 2\left( {{x^2} – 3x} \right) \cr & Q = 2\left( {{x^2} – 2.x.{3 \over 2} + {9 \over 4}} \right) – {9 \over 2} \cr & Q = 2{\left( {x – {3 \over 2}} \right)^2} – {9 \over 2} \cr & \Rightarrow Q \ge – {9 \over 2} \Rightarrow {Q_{\min }} = – {9 \over 2} \cr & Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x = {3 \over 2} \cr & c)\,\,M = {x^2} + {y^2} – x + 6y + 10 \cr & M = \left( {{x^2} – x + {1 \over 4}} \right) + \left( {{y^2} + 6y + 9} \right) + {3 \over 4} \cr & M = {\left( {x – {1 \over 2}} \right)^2} + {\left( {y + 3} \right)^2} + {3 \over 4} \cr & \Rightarrow M \ge {3 \over 4} \Rightarrow {M_{\min }} = {3 \over 4} \cr & Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \left\{ \matrix{ x = {1 \over 2} \hfill \cr y = – 3 \hfill \cr} \right. \cr} $$ Bình luận
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 Bình luận
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:
$$\eqalign{
& a)\,\,P = {x^2} – 2x + 5 \cr
& P = {x^2} – 2x + 1 + 4 \cr
& P = {\left( {x – 1} \right)^2} + 4 \cr
& P \ge 4 \Rightarrow {P_{\min }} = 4 \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1 \cr
& b)\,\,Q = 2{x^2} – 6x \cr
& Q = 2\left( {{x^2} – 3x} \right) \cr
& Q = 2\left( {{x^2} – 2.x.{3 \over 2} + {9 \over 4}} \right) – {9 \over 2} \cr
& Q = 2{\left( {x – {3 \over 2}} \right)^2} – {9 \over 2} \cr
& \Rightarrow Q \ge – {9 \over 2} \Rightarrow {Q_{\min }} = – {9 \over 2} \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x = {3 \over 2} \cr
& c)\,\,M = {x^2} + {y^2} – x + 6y + 10 \cr
& M = \left( {{x^2} – x + {1 \over 4}} \right) + \left( {{y^2} + 6y + 9} \right) + {3 \over 4} \cr
& M = {\left( {x – {1 \over 2}} \right)^2} + {\left( {y + 3} \right)^2} + {3 \over 4} \cr
& \Rightarrow M \ge {3 \over 4} \Rightarrow {M_{\min }} = {3 \over 4} \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \left\{ \matrix{
x = {1 \over 2} \hfill \cr
y = – 3 \hfill \cr} \right. \cr} $$
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