Find the maximum value of:
a) \(-3x^2-9x-25\)
b) \(x-x^2\)
c) \(-x^2+7x+12\)
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20 tháng 7 2022
a: \(A=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7< =7\)
Dấu '=' xảy ra khi x=2
b: \(B=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=1/2
c: \(C=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
e: \(E=-\left(x^2+6x+9+1\right)=-\left(x+3\right)^2-1< =-1\)
Dấu = xảy ra khi x=-3
LM
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L
31 tháng 8 2018
a) A = \(2x^2+x-1=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)\)\(-\frac{9}{8}=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\)
Vì \(\left(x+\frac{1}{4}\right)^2\ge0\forall x\Leftrightarrow2\left(x+\frac{1}{4}\right)^2\ge0\forall x\Leftrightarrow2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\forall x\Leftrightarrow A\ge-\frac{9}{8}\)
Dấu = xảy ra \(\Leftrightarrow\)\(x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)
Vậy minA =\(-\frac{9}{8}\)khi \(x=-\frac{1}{4}\).
b) B=\(5x-3x^2+2=-3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)+\frac{49}{12}=-3\left(x-\frac{5}{6}\right)^2+\frac{49}{12}\)
Vì \(\left(x-\frac{5}{6}\right)^2\ge0\forall x\Leftrightarrow-3\left(x-\frac{5}{6}\right)^2\le0\forall x\Leftrightarrow-3\left(x-\frac{5}{6}\right)^2+\frac{49}{12}\le\frac{49}{12}\forall x\Leftrightarrow B\le\frac{49}{12}\forall x\)
Dấu = xảy ra \(\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)
Vậy maxB = \(\frac{49}{12}\)khi \(x=\frac{5}{6}\).
DP
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15 tháng 8 2019
a) \(x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
15 tháng 8 2019
b) \(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
20 tháng 6 2022
a: \(\Leftrightarrow\left(3x+2\right)\left(5-x\right)=-9x^2+4\)
\(\Leftrightarrow\left(3x+2\right)\left(5-x\right)+\left(3x+2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(2x+3\right)=0\)
=>x=-2/3 hoặc x=-3/2
b: \(\Leftrightarrow4x\left(x+5\right)+x^2-25=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x-5\right)=0\)
=>x=-5 hoặc x=1
c: \(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
=>x=1 hoặc x=-1/2
a: \(=-3\left(x^2+3x+\dfrac{25}{3}\right)\)
\(=-3\left(x^2+3x+\dfrac{9}{4}+\dfrac{73}{12}\right)\)
\(=-3\left(x+\dfrac{3}{2}\right)^2-\dfrac{73}{4}< =-\dfrac{73}{4}\)
Dấu '=' xảy ra khi x=-3/2
b: \(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=1/2
c: \(=-\left(x^2-7x-12\right)\)
\(=-\left(x^2-7x+\dfrac{49}{4}-\dfrac{97}{4}\right)\)
\(=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{97}{4}< =\dfrac{97}{4}\)
Dấu '=' xảy ra khi x=7/2