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\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)
a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Dấu "=" \(\Leftrightarrow x=-1\)
b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)
c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)
Dấu "=" \(\Leftrightarrow x=2\)
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\(A=-2x^2+2\left|x\right|-5=-\dfrac{1}{2}\left(4x^2-4\left|x\right|+1\right)-\dfrac{9}{2}=-\dfrac{1}{2}\left(2\left|x\right|-1\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)
\(A_{max}=-\dfrac{9}{2}\) khi \(\left|x\right|=\dfrac{1}{2}\Rightarrow x=\pm\dfrac{1}{2}\)
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1) \(M=9x^2-6x+6=\left(9x^2-6x+1\right)+5=\left(3x-1\right)^2+5\ge5\)
\(minM=5\Leftrightarrow x=\dfrac{1}{3}\)
2) \(M=5-2x-x^2=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)
\(maxM=6\Leftrightarrow x=-1\)
3) \(N=5+6x-9x^2=-\left(9x^2-6x+1\right)+6=-\left(3x-1\right)^2+6\le6\)
\(maxN=6\Leftrightarrow x=\dfrac{1}{3}\)
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\(x^2+2x+5\)
\(=\left(x+1\right)^2+4\ge4\forall x\)
\(\Leftrightarrow\dfrac{5}{x^2+2x+5}\le\dfrac{5}{4}\forall x\)
Dấu '=' xảy ra khi x=-1
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c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)
\(\Leftrightarrow V\ge-1\forall x\)
Dấu '=' xảy ra khi x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-2x^2+2\left|x\right|-5\)
\(=-2\left(x^2-\left|x\right|\right)-5\)
\(=-2\left(\left|x\right|^2-2\left|x\right|\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}-5\)
\(=-2\left(\left|x\right|-\frac{1}{2}\right)^2-\frac{19}{5}\)
Mà: \(\left(\left|x\right|-\frac{1}{2}\right)^2\ge0\)
\(\Leftrightarrow-2\left(\left|x\right|-\frac{1}{2}\right)^2\le0\)
\(\Leftrightarrow-2\left(\left|x\right|-\frac{1}{2}\right)^2-\frac{19}{5}\le\frac{-19}{5}\)
Dấu '' = '' xảy ra \(\Leftrightarrow\left|x\right|-\frac{1}{2}=0\Leftrightarrow\left|x\right|=\frac{1}{2}\Leftrightarrow x=\pm\frac{1}{2}\)