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\(P=\left(x+3\right)^2+y^2+5\ge5\)
\(P_{min}=5\) khi \(\left\{{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+5\)
\(=\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)+5\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)+5\)
\(=\left(x^2-5x+4\right)^2-2\left(x^2-5x+4\right)+1+4\)
\(=\left(x^2-5x+4-1\right)^2+4\)
\(=\left(x^2-5x+3\right)^2+4\)
Vì \(\left(x^2-5x+3\right)^2\ge0\Rightarrow\left(x^2-5x+3\right)^2+4\ge4\)
Vậy giá trị nhỏ nhất của \(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+5\) là \(4\)
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*\(x\ge\dfrac{1}{2}\Leftrightarrow\left|2x-1\right|=2x-1\)
\(D=\left(2x-1\right)^2-3\left(2x-1\right)+2=\left(2x-1\right)^2-2.\dfrac{3}{2}\left(2x-1\right)+\dfrac{9}{4}-\dfrac{1}{4}=\left(2x-1-\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(2x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)\(D_{min}=-\dfrac{1}{4}\Leftrightarrow x=\dfrac{5}{4}\left(1\right)\)
*\(x< \dfrac{1}{2}\Leftrightarrow\left|2x-1\right|=-2x+1\)
\(D=\left(2x-1\right)^2+3\left(2x-1\right)+2=\left(2x-1\right)^2+2.\dfrac{3}{2}\left(2x-1\right)+\dfrac{9}{4}-\dfrac{1}{4}=\left(2x-1+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(2x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)\(D_{min}=-\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\left(2\right)\)
-Từ (1) và (2) suy ra \(D_{min}=-\dfrac{1}{4}\Leftrightarrow x\in\left\{\dfrac{5}{4};\dfrac{-1}{4}\right\}\)
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\(A=3x-x^2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{9}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)
Vậy GTLN của A là \(\frac{9}{4}\)khi x = \(\frac{3}{2}\)
\(B=7-8x-x^2=-\left(x^2+8x+16\right)+23=-\left(x+4\right)^2+23\le23\)
Vậy GTLN của B là 23 khi x = -4
\(C=x^2-20x+101=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)
Vậy GTNN của C là 1 khi x = 10
\(D=3x^2-6x+11=3\left(x^2-2x+1\right)+8=3\left(x-1\right)^2+8\ge8\)
Vậy GTNN của D là 8 khi x = 1
\(a,A=3x-x^2=-x^2+3x=-x^2+2.\frac{3}{2}x-\frac{9}{4}+\frac{9}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)
Vậy Max A = 9/4 <=> x = 3/2
\(b,B=7-8x-x^2=-x^2-8x+7=-x^2-2.4x-16+23=-\left(x+4\right)^2+23\ge23\)
Vậy MinB = 23 <=> x = -4
\(c,C=x^2-20x+101=x^2-2.10x+10^2+1=\left(x-10\right)^2+1\ge1\)
Vậy MinC = 1 <=> x = 10
\(d,D=3x^2-6x+11\)
\(D=\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\sqrt{3}+\left(\sqrt{3}\right)^2+8=\left(\sqrt{3}x-\sqrt{3}\right)^2+8\ge8\)
Vậy MinD = 8<=> x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=a^2+b^2=\frac{a^2}{1}+\frac{b^2}{1}\ge\frac{\left(a+b\right)^2}{2}=8\)
Vậy \(Min_A=8\)
Khi \(a=b=2\)
x2 - 5x= (x2- 2.x.\(\frac{5}{2}\)+ \(\frac{25}{4}\) )-\(\frac{25}{4}\)
=(x-\(\frac{5}{2}\))2 -\(\frac{25}{4}\) \(\le\) \(\frac{25}{4}\)
vậy giá trị nhỏ nhất cua x2- 5x là \(\frac{25}{4}\) tại x =\(\frac{5}{2}\)
cảm ơn bạn nhìu