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\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)
a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)
\(minA=2\Leftrightarrow x=3\)
b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)
\(minB=51\Leftrightarrow x=5\)
c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
a)
Ta có:
\(A=x^2-2x-1=x^2-2x+1-2=\left(x-1\right)^2-2\)
\(\ge0-2=-2\)
Vậy \(A_{min}=-2\), đạt được khi và chỉ khi \(x-1=0\Leftrightarrow x=1\)
b)\(B=4x^2+4x+8=4x^2+4x+1+7\)
\(=\left(2x+1\right)^2+7\ge0+7=7\)
Vậy \(B_{min}=7\), đạt được khi và chỉ khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)
c)
Ta có:
\(C=3x-x^2+2=2-\left(x^2-3x\right)\)
\(=2+\dfrac{9}{4}-\left(x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}\right)\)
\(=\dfrac{17}{4}-\left(x-\dfrac{3}{2}\right)^2\le\dfrac{17}{4}-0=\dfrac{17}{4}\)
Vậy \(C_{max}=\dfrac{17}{4}\), đạt được khi và chỉ khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
d) Ta có:
\(D=-x^2-5x=-\left(x^2+5x\right)=\dfrac{25}{4}-\left(x^2+2x.\dfrac{5}{2}+\dfrac{25}{4}\right)\)
\(=\dfrac{25}{4}-\left(x+\dfrac{5}{2}\right)^2\le\dfrac{25}{4}-0=\dfrac{25}{4}\)
Vậy \(D_{max}=\dfrac{25}{4}\), đạt được khi và chỉ khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)
e) Ta có:
\(E=x^2-4xy+5y^2+10x-22y+28\)
\(=x^2+4y^2+5^2-4xy+10x-20y+y^2-2y+1+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)
\(\ge0+0+2=2\)
Vậy \(E_{min}=2\), đạt được khi và chỉ khi \(x-2y+5=y-1=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
\(a,A=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ A=\left(x-2y\right)^2+10\left(x-2y\right)+5+\left(y-1\right)^2+2\\ A=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=2y-5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
\(b,\Leftrightarrow3x^3+10x^2-5+n=\left(3x+1\right)\cdot a\left(x\right)\)
Thay \(x=-\dfrac{1}{3}\Leftrightarrow3\left(-\dfrac{1}{27}\right)+10\cdot\dfrac{1}{9}-5+n=0\)
\(\Leftrightarrow-\dfrac{1}{9}+\dfrac{10}{9}-5+n=0\\ \Leftrightarrow-4+n=0\Leftrightarrow n=4\)
\(c,\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\\ \Leftrightarrow2n\left(n-2\right)+5\left(n-2\right)+3⋮n-2\\ \Leftrightarrow n-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow n\in\left\{-1;1;3;5\right\}\)
a) \(M=x^2-3x+10\)
\(M=x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}+\dfrac{31}{4}\)
\(M=\left(x^2-2\cdot\dfrac{3}{2}\cdot x+\dfrac{9}{4}\right)+\dfrac{31}{4}\)
\(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\)
Mà: \(\left(x-\dfrac{3}{2}\right)^2\ge0\) nên: \(M=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)
Dấu "=" xảy ra
\(\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=\dfrac{31}{4}\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(M_{min}=\dfrac{31}{4}\) với \(x=\dfrac{3}{2}\)
b) \(N=2x^2+5y^2+4xy+8x-4y-100\)
\(N=x^2+x^2+4y^2+y^2+4xy+8x-4y-120+16+4\)
\(N=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-120\)
\(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\)
Mà:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\\\left(x+4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) nên \(N=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-120\ge120\)
Dấu "=" xảy ra:
\(\left\{{}\begin{matrix}\left(x+2y\right)^2=0\\\left(x+4\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-4+2y=0\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-4\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Vậy: \(N_{min}=120\) khi \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
b) Ta có : 4x - x2 + 1
= -(x2 - 4x - 1)
= -(x2 - 4x + 4 - 5)
= -(x2 - 4x + 4) + 5
= -(x - 2)2 + 5 \(\le5\forall x\) vì : \(-\left(x-2\right)^2\le0\forall x\)
Vậy GTLN của biểu thức là : 5 khi x = 2
Ta có : (x2 - 4xy + 4y2) + (10x - 20y) + (y2 - 2y + 1) + 27
= (x - 2y)2 + 10(x - 2y) + (y - 1)2
= (x - 2y)2 + 10(x - 2y) + 25 + (y - 1)2 + 2
= (x - 2y + 5)2 + (y - 1)2 + 2 \(\ge2\forall x\)
Vậy GTNN của biểu thức là 2
Khi \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
a) \(A=4x^2+7x+13=4\left(x^2+\frac{7}{4}x+\frac{49}{64}\right)+\frac{159}{16}\)
\(=4\left(x+\frac{7}{8}\right)^2+\frac{159}{16}\ge\frac{159}{16}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(4\left(x+\frac{7}{8}\right)^2=0\Rightarrow x=-\frac{7}{8}\)
Vậy \(A_{Min}=\frac{159}{16}\Leftrightarrow x=-\frac{7}{8}\)
b) \(B=5-8x+x^2=\left(x^2-8x+16\right)-11\)
\(=\left(x-4\right)^2-11\ge-11\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-4\right)^2=0\Rightarrow x=4\)
Vậy \(B_{Min}=-11\Leftrightarrow x=4\)