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\(B=2x\left(x-4\right)-10=2x^2-8x-10\)
\(=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)
\(minB=-18\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(N=-1-x-x^2=-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\)
\(maxN=-\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)
\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)
a: Ta có: \(N=-x^2-x-1\)
\(=-\left(x^2+x+1\right)\)
\(=-\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
b: ta có: \(B=3x^2+4x-13\)
\(=3\left(x^2+\dfrac{4}{3}x-\dfrac{13}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{43}{9}\right)\)
\(=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{43}{3}\ge-\dfrac{43}{3}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{2}{3}\)
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a) \(A=2x^2\)\(+\)\(10\)\(-\)\(1\)
\(=2\left(x^2+5x-\frac{1}{2}\right)\)
\(=2\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(=2\left(x+\frac{5}{2}\right)^2\)\(=\frac{27}{2}\)> hoặc = \(\frac{-27}{2}\)\(=-13,5\)
Dấu bằng xảy ra \(\Leftrightarrow\)\(x+\frac{5}{2}=0\)
\(x=\frac{-5}{2}=-2,5\)
Vậy GTLN của A bằng -13,5 khi x = -2,5
b) \(B=3x-2x^2\)
\(=\)\(-2\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}-\frac{9}{16}\right)\)
\(=-2\left[\left(x-\frac{3}{4}\right)^2-\frac{9}{16}\right]\)
\(=-2\left(x-0,75\right)^2\)\(+\)\(\frac{9}{8}\)< hoặc = \(\frac{9}{8}\)\(=\)\(1,125\)
Dấu bằng xảy ra \(\Leftrightarrow\)\(x-0,75=0\)
\(x=0,75\)
Vậy GTLN của B bằng 1,125 khi x = 0,75
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2-2x=3\left(x^2-\frac{2}{3}x\right)=3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}-\frac{1}{9}\right)\)
\(=3\left[\left(x-\frac{1}{3}\right)^2-\frac{1}{9}\right]=3\left(x-\frac{1}{3}\right)^2-\frac{1}{3}\ge\frac{-1}{3}\)
Vậy GTNN của bt là \(\frac{-1}{3}\Leftrightarrow x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\)
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\(A=2x^2-3x+2=2\left(x^2-\frac{3}{2}x\right)+2\)
\(=2\left(x^2-2.\frac{3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+2=2\left(x-\frac{3}{4}\right)^2-\frac{9}{8}+2\ge\frac{7}{8}\)
Dấu ''='' xảy ra khi x = 3/4
Vậy GTNN của A bằng 7/8 tại x = 3/4
![](https://rs.olm.vn/images/avt/0.png?1311)