Tìm GTNN: M= -3x^2-x+5 N=3x+x^2-2 H=3x^2-y+2y^2+x-12 giúp mik với hứa 5 sao và CTLHN nếu đúng 04/07/2021 Bởi Cora Tìm GTNN: M= -3x^2-x+5 N=3x+x^2-2 H=3x^2-y+2y^2+x-12 giúp mik với hứa 5 sao và CTLHN nếu đúng
Đáp án: Không tồn tại x để M đạt GTNN \(MinN = – \dfrac{{17}}{4}\) \(MinH = – \dfrac{{293}}{{24}}\) Giải thích các bước giải: \(\begin{array}{l}M = – 3{x^2} – x + 5\\ = – \left( {3{x^2} + x – 5} \right)\\ = – \left( {3{x^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}} – \dfrac{{61}}{{12}}} \right)\\ = – {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{61}}{{12}}\\Do:{\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \ge 0\forall x\\ \to – {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \le 0\\ \to – {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{61}}{{12}} \le \dfrac{{61}}{{12}}\\ \to Max = \dfrac{{61}}{{12}}\\ \Leftrightarrow x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }} = 0\\ \to x = – \dfrac{1}{6}\end{array}\) ⇒ Không tồn tại x để M đạt GTNN \(\begin{array}{l}N = {x^2} + 3x – 2\\ = {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} – \dfrac{{17}}{4}\\ = {\left( {x + \dfrac{3}{2}} \right)^2} – \dfrac{{17}}{4}\\Do:{\left( {x + \dfrac{3}{2}} \right)^2} \ge 0\forall x\\ \to {\left( {x + \dfrac{3}{2}} \right)^2} – \dfrac{{17}}{4} \ge – \dfrac{{17}}{4}\\ \to Min = – \dfrac{{17}}{4}\\ \Leftrightarrow x = – \dfrac{3}{2}\\H = 3{x^2} – y + 2{y^2} + x – 12\\ = 3{x^2} + x + 2{y^2} – y – 12\\ = \left( {3{x^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}}} \right) + \left( {2{y^2} – 2.y\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8}} \right) – \dfrac{{293}}{{24}}\\ = {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2 – \dfrac{1}{{2\sqrt 2 }}} \right)^2} – \dfrac{{293}}{{24}}\\Do:{\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2 – \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x;y\\ \to {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2 – \dfrac{1}{{2\sqrt 2 }}} \right)^2} – \dfrac{{293}}{{24}} \ge – \dfrac{{293}}{{24}}\\ \to Min = – \dfrac{{293}}{{24}}\\ \Leftrightarrow \left\{ \begin{array}{l}x = – \dfrac{1}{6}\\y = – \dfrac{1}{4}\end{array} \right.\end{array}\) Bình luận
Đáp án: `M=-3x^2-x+5` `=-3(x^2+1/3x)+5` `=-3(x^2+1/3x+1/36-1/36)+5` `=-3(x+1/6)^2+1/12+5` `=-3(x+1/6)^2+61/12<=61/12` Dấu “=” xảy ra khi `x+1/6=0<=>x=-1/6` `N=3x+x^2-2` `=x^2+3x-2` `=x^2+3x+9/4-9/4-2` `=(x+3/2)^2-17/4>=-17/4` Dấu “=” xảy ra khi `x=-3/2` `H=3x^2-y+2y^2+x-12` `=3(x^2+1/3x)+2(y^2-1/2y)-12` `=3(x^2+1/3x+1/36-1/36)+2(y^2-1/2y+1/16-1/16)-12` `=3(x+1/6)^2+2(y-1/4)^2-1/12-1/8-12` `=3(x+1/6)^2+2(y-1/4)^2-293/24>=-293/24` Dấu “=” xảy ra khi \(\begin{cases}x+\dfrac16=0\\y-\dfrac14=0\\\end{cases}\) `<=>` \(\begin{cases}x=-\dfrac16\\y=\dfrac14\\\end{cases}\) Bình luận
Đáp án:
Không tồn tại x để M đạt GTNN
\(MinN = – \dfrac{{17}}{4}\)
\(MinH = – \dfrac{{293}}{{24}}\)
Giải thích các bước giải:
\(\begin{array}{l}
M = – 3{x^2} – x + 5\\
= – \left( {3{x^2} + x – 5} \right)\\
= – \left( {3{x^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}} – \dfrac{{61}}{{12}}} \right)\\
= – {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{61}}{{12}}\\
Do:{\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \ge 0\forall x\\
\to – {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \le 0\\
\to – {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{61}}{{12}} \le \dfrac{{61}}{{12}}\\
\to Max = \dfrac{{61}}{{12}}\\
\Leftrightarrow x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }} = 0\\
\to x = – \dfrac{1}{6}
\end{array}\)
⇒ Không tồn tại x để M đạt GTNN
\(\begin{array}{l}
N = {x^2} + 3x – 2\\
= {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} – \dfrac{{17}}{4}\\
= {\left( {x + \dfrac{3}{2}} \right)^2} – \dfrac{{17}}{4}\\
Do:{\left( {x + \dfrac{3}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x + \dfrac{3}{2}} \right)^2} – \dfrac{{17}}{4} \ge – \dfrac{{17}}{4}\\
\to Min = – \dfrac{{17}}{4}\\
\Leftrightarrow x = – \dfrac{3}{2}\\
H = 3{x^2} – y + 2{y^2} + x – 12\\
= 3{x^2} + x + 2{y^2} – y – 12\\
= \left( {3{x^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}}} \right) + \left( {2{y^2} – 2.y\sqrt 2 .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8}} \right) – \dfrac{{293}}{{24}}\\
= {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2 – \dfrac{1}{{2\sqrt 2 }}} \right)^2} – \dfrac{{293}}{{24}}\\
Do:{\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2 – \dfrac{1}{{2\sqrt 2 }}} \right)^2} \ge 0\forall x;y\\
\to {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + {\left( {y\sqrt 2 – \dfrac{1}{{2\sqrt 2 }}} \right)^2} – \dfrac{{293}}{{24}} \ge – \dfrac{{293}}{{24}}\\
\to Min = – \dfrac{{293}}{{24}}\\
\Leftrightarrow \left\{ \begin{array}{l}
x = – \dfrac{1}{6}\\
y = – \dfrac{1}{4}
\end{array} \right.
\end{array}\)
Đáp án:
`M=-3x^2-x+5`
`=-3(x^2+1/3x)+5`
`=-3(x^2+1/3x+1/36-1/36)+5`
`=-3(x+1/6)^2+1/12+5`
`=-3(x+1/6)^2+61/12<=61/12`
Dấu “=” xảy ra khi `x+1/6=0<=>x=-1/6`
`N=3x+x^2-2`
`=x^2+3x-2`
`=x^2+3x+9/4-9/4-2`
`=(x+3/2)^2-17/4>=-17/4`
Dấu “=” xảy ra khi `x=-3/2`
`H=3x^2-y+2y^2+x-12`
`=3(x^2+1/3x)+2(y^2-1/2y)-12`
`=3(x^2+1/3x+1/36-1/36)+2(y^2-1/2y+1/16-1/16)-12`
`=3(x+1/6)^2+2(y-1/4)^2-1/12-1/8-12`
`=3(x+1/6)^2+2(y-1/4)^2-293/24>=-293/24`
Dấu “=” xảy ra khi \(\begin{cases}x+\dfrac16=0\\y-\dfrac14=0\\\end{cases}\)
`<=>` \(\begin{cases}x=-\dfrac16\\y=\dfrac14\\\end{cases}\)