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Ta có: \(P=2x-2xy-2x^2-y^2\)
\(P=-x^2-2xy-y^2-x^2+2x\)
\(P=-\left(x^2+2xy+y^2\right)-\left(x^2-2x+1\right)+1\)
\(P=-\left(x+y\right)^2-\left(x-1\right)^2+1\)
\(P=-\left[\left(x+y\right)^2+\left(x-1\right)^2\right]+1\le1\forall x;y\)
Vậy GTLN của P là 1 khi x=-1; y=1.
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b: \(-7x^2-4y^2-8xy+18x+9\)
\(=\left(-4x^2-8xy-4y^2\right)+\left(-3x^2+18x-27\right)+36\)
\(=-4\left(x^2+2xy+y^2\right)-3\left(x^2-6x+9\right)+36\)
\(=-4\left(x+y\right)^2-3\left(x-3\right)^2+36< =36\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\x-3=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)
a.
Đặt \(A=-x^2-5y^2+4xy+12x+7\)
\(=-\dfrac{1}{5}\left(4x^2-20xy+25y^2\right)-\dfrac{1}{5}\left(x^2-60x+900\right)+187\)
\(=-\dfrac{1}{5}\left(2x-5y\right)^2-\dfrac{1}{5}\left(x-30\right)^2+187\le187\)
\(A_{max}=187\) khi \(\left(x;y\right)=\left(30;12\right)\)
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a) \(A=6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)
Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).
b) \(B=x^2-5x-2=x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2-\frac{33}{4}=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge-\frac{33}{44}\)
Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).
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Câu này em đã hỏi rồi
1.Tìm GTNN của Bthức : B= 4x2- 6x+1 : (x-2)2 với x ≠ 22. Tìm GTLN của Bthức: C= x2 + 4x - 14 : x2 -2x +1 với x≠ 1gi... - Hoc24
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\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)
Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).
\(B=\frac{x^2}{x^2+x+1}=\frac{3x^2}{3x^2+3x+3}=\frac{4x^2+4x+4-\left(x^2+4x+4\right)}{3x^2+3x+3}=\frac{4}{3}-\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\le\frac{4}{3}\)
Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).
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Lời giải:
$K=-5x^2+20x-2021=-2001-5(x^2-4x+4)=-2001-5(x-2)^2$
Vì $(x-2)^2\geq 0, \forall x\in\mathbb{R}$
$\Rightarrow K=-2001-5(x-2)^2\leq -2001$
Vậy $K_{\max}=-2001$ khi $(x-2)^2=0\Leftrightarrow x=2$
Ta có: \(K=-5x^2+20x-2021\)
\(=-5\left(x^2-4x+\dfrac{2021}{5}\right)\)
\(=-5\left(x^2-4x+4+\dfrac{2001}{5}\right)\)
\(=-5\left(x-2\right)^2-2001\le-2001\forall x\)
Dấu '=' xảy ra khi x=2
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Ta có: \(-x^2+4x+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\forall x\in R\)
⇒ max y = 7 tại x = 2
Đặt \(A=-3x^2-5y^2+2x+7y-23\)
\(=-3\left(x^2-\dfrac{2}{3}x\right)-5\left(y^2-\dfrac{7}{5}y\right)-23\)
\(=-3\left(x^2-2.\dfrac{1}{3}x+\dfrac{1}{9}-\dfrac{1}{9}\right)-5\left(y^2-2.\dfrac{7}{10}y+\dfrac{49}{100}-\dfrac{49}{100}\right)-23\)
\(=-3\left(x^2-2.\dfrac{1}{3}x+\dfrac{1}{9}\right)+\dfrac{1}{3}-5\left(y^2-2.\dfrac{7}{10}y+\dfrac{49}{100}\right)+\dfrac{49}{20}-23\)
\(=-3\left(x-\dfrac{1}{3}\right)^2-5\left(y-\dfrac{7}{10}\right)^2-\dfrac{1213}{60}\le-\dfrac{1213}{60}\)
Vậy \(A_{max}=-\dfrac{1213}{60}\) khi \(x=\dfrac{1}{3};y=\dfrac{7}{10}\)