Tìm GTLN,GTNN 1)A=x^2-2x+5 2)B=4x-x^2+3 3)C=x^2+y^2-x+6y+10 4)D=5.x-x^2 5)E=x^2+10y^2-6xy-2y+3 17/09/2021 Bởi Emery Tìm GTLN,GTNN 1)A=x^2-2x+5 2)B=4x-x^2+3 3)C=x^2+y^2-x+6y+10 4)D=5.x-x^2 5)E=x^2+10y^2-6xy-2y+3
Đáp án: Giải thích các bước giải: 1, A= (x^2-2x+1)+4=(x-1)^2+4 Vì ${\left( {x – 1} \right)^2} \ge 0\forall x$ =>${\left( {x – 1} \right)^2} + 4 \ge 4\forall x$ => MinA=4 khi x=1 2, B= -(x^2-4x+4)+7=-${\left( {x – 2} \right)^2}$+7 -${\left( {x – 2} \right)^2}$$ \le $0 với mọi x => -${\left( {x – 2} \right)^2}$+7 <=7 với mọi x => Max B=7 khi x=2 3, C=(${x^2} – x + 1/4) + ({y^2} + 6y + 9)$=3/4 =${\left( {x – 1/2} \right)^2} + {(y + 3)^2} + 3/4$ $ \ge 3/4\forall x,y \in R$ => Min C = 3/4 khi x=1/2, y=-3 4, D= -(x^2-5x+25/4)+25/4=-(x-5/2)^2+25/4 $ \le $ 25/4 với mọi x => MaxD=25/4 5, (x^2 -6xy+9y^2)+(y^2-2y+1)+2=(x-3y)^2+(y-1)^2+2 $ \ge $ với mọi x,y thuộc R => MinE=2 Bình luận
\(\begin{array}{l} 1)\,\,\,A = {x^2} – 2x + 5 = {\left( {x – 1} \right)^2} + 4 \ge 4\\ Dau\,\, = \,\,xay\,\,\,ra \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1.\\ Vay\,\,Min\,\,A = 4\,\,khi\,\,x = 1.\\ 2)\,\,\,B = 4x – {x^2} + 3 = – \left( {{x^2} – 4x + 4} \right) + 4 + 3\\ = – {\left( {x – 2} \right)^2} + 7 \le 7\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x – 2 = 0 \Leftrightarrow x = 2\\ Vay\,\,Max\,\,B = 7\,\,khi\,\,\,x = 2.\\ 3)\,\,C = {x^2} + {y^2} – x + 6y + 10 = {x^2} – 2.\frac{1}{2}x + \frac{1}{4} + \left( {{y^2} + 6y + 9} \right) + \frac{3}{4}\\ = {\left( {x – \frac{1}{2}} \right)^2} + {\left( {y + 3} \right)^2} + \frac{3}{4} \ge \frac{3}{4}\\ Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow \left\{ \begin{array}{l} x – \frac{1}{2} = 0\\ y + 3 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \frac{1}{2}\\ y = – 3 \end{array} \right.\\ Vay\,\,Min\,C = \frac{3}{4}\,\,\,khi\,\,\,x = \frac{1}{2};\,\,\,y = – 3.\\ 4)\,\,D = 5x – {x^2} = – \left( {{x^2} – 2.\frac{5}{2}x + \frac{{25}}{4}} \right) + \frac{{25}}{4}\\ = – {\left( {x – \frac{5}{2}} \right)^2} + \frac{{25}}{4} \le \frac{{25}}{4}\\ Dau\,\,\, = \,\,\,xay\,\,ra \Leftrightarrow x – \frac{5}{2} = 0 \Leftrightarrow x = \frac{5}{2}\\ Vay\,\,MaxD = \frac{{25}}{4}\,\,\,khi\,\,\,x = \frac{5}{2}.\\ 5)\,\,E = {x^2} + 10{y^2} – 6xy – 2y + 3 = {x^2} – 6xy + 9{y^2} + {y^2} – 2y + 1 + 2\\ = {\left( {x – 3y} \right)^2} + {\left( {y – 1} \right)^2} + 2 \ge 2\\ Dau\,\, = \,\,xay\,\,ra\, \Leftrightarrow \left\{ \begin{array}{l} x – 3y = 0\\ y – 1 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 3\\ y = 1 \end{array} \right.\\ Vay\,\,Min\,E = 2\,\,\,khi\,\,x = 3;\,\,y = 1. \end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
1, A= (x^2-2x+1)+4=(x-1)^2+4
Vì ${\left( {x – 1} \right)^2} \ge 0\forall x$
=>${\left( {x – 1} \right)^2} + 4 \ge 4\forall x$
=> MinA=4 khi x=1
2, B= -(x^2-4x+4)+7=-${\left( {x – 2} \right)^2}$+7
-${\left( {x – 2} \right)^2}$$ \le $0 với mọi x
=> -${\left( {x – 2} \right)^2}$+7 <=7 với mọi x
=> Max B=7 khi x=2
3, C=(${x^2} – x + 1/4) + ({y^2} + 6y + 9)$=3/4
=${\left( {x – 1/2} \right)^2} + {(y + 3)^2} + 3/4$ $ \ge 3/4\forall x,y \in R$
=> Min C = 3/4 khi x=1/2, y=-3
4, D= -(x^2-5x+25/4)+25/4=-(x-5/2)^2+25/4 $ \le $ 25/4 với mọi x
=> MaxD=25/4
5, (x^2 -6xy+9y^2)+(y^2-2y+1)+2=(x-3y)^2+(y-1)^2+2 $ \ge $ với mọi x,y thuộc R
=> MinE=2
\(\begin{array}{l}
1)\,\,\,A = {x^2} – 2x + 5 = {\left( {x – 1} \right)^2} + 4 \ge 4\\
Dau\,\, = \,\,xay\,\,\,ra \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1.\\
Vay\,\,Min\,\,A = 4\,\,khi\,\,x = 1.\\
2)\,\,\,B = 4x – {x^2} + 3 = – \left( {{x^2} – 4x + 4} \right) + 4 + 3\\
= – {\left( {x – 2} \right)^2} + 7 \le 7\\
Dau\,\, = \,\,xay\,\,ra \Leftrightarrow x – 2 = 0 \Leftrightarrow x = 2\\
Vay\,\,Max\,\,B = 7\,\,khi\,\,\,x = 2.\\
3)\,\,C = {x^2} + {y^2} – x + 6y + 10 = {x^2} – 2.\frac{1}{2}x + \frac{1}{4} + \left( {{y^2} + 6y + 9} \right) + \frac{3}{4}\\
= {\left( {x – \frac{1}{2}} \right)^2} + {\left( {y + 3} \right)^2} + \frac{3}{4} \ge \frac{3}{4}\\
Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow \left\{ \begin{array}{l}
x – \frac{1}{2} = 0\\
y + 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{1}{2}\\
y = – 3
\end{array} \right.\\
Vay\,\,Min\,C = \frac{3}{4}\,\,\,khi\,\,\,x = \frac{1}{2};\,\,\,y = – 3.\\
4)\,\,D = 5x – {x^2} = – \left( {{x^2} – 2.\frac{5}{2}x + \frac{{25}}{4}} \right) + \frac{{25}}{4}\\
= – {\left( {x – \frac{5}{2}} \right)^2} + \frac{{25}}{4} \le \frac{{25}}{4}\\
Dau\,\,\, = \,\,\,xay\,\,ra \Leftrightarrow x – \frac{5}{2} = 0 \Leftrightarrow x = \frac{5}{2}\\
Vay\,\,MaxD = \frac{{25}}{4}\,\,\,khi\,\,\,x = \frac{5}{2}.\\
5)\,\,E = {x^2} + 10{y^2} – 6xy – 2y + 3 = {x^2} – 6xy + 9{y^2} + {y^2} – 2y + 1 + 2\\
= {\left( {x – 3y} \right)^2} + {\left( {y – 1} \right)^2} + 2 \ge 2\\
Dau\,\, = \,\,xay\,\,ra\, \Leftrightarrow \left\{ \begin{array}{l}
x – 3y = 0\\
y – 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 1
\end{array} \right.\\
Vay\,\,Min\,E = 2\,\,\,khi\,\,x = 3;\,\,y = 1.
\end{array}\)