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\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
\(a,A=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ A=\left(x-2y\right)^2+10\left(x-2y\right)+5+\left(y-1\right)^2+2\\ A=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
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 \(x=-\dfrac{1}{3}\Leftrightarrow3\left(-\dfrac{1}{27}\right)+10\cdot\dfrac{1}{9}-5+n=0\)
\(\Leftrightarrow-\dfrac{1}{9}+\dfrac{10}{9}-5+n=0\\ \Leftrightarrow-4+n=0\Leftrightarrow n=4\)
\(c,\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\\ \Leftrightarrow2n\left(n-2\right)+5\left(n-2\right)+3⋮n-2\\ \Leftrightarrow n-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow n\in\left\{-1;1;3;5\right\}\)
\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)
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) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
A = x2 - 4xy + 5y2 + 10x - 22y + 2044
= ( x2 - 4xy + 4y2 + 10x - 20y + 25 ) + ( y2 - 2y + 1 ) + 2018
= [ ( x2 - 4xy + 4y2 ) + ( 10x - 20y ) + 25 ] + ( y - 1 )2 + 2018
= [ ( x - 2y )2 + 2( x - 2y ).5 + 52 ] + ( y - 1 )2 + 2018
= ( x - 2y + 5 )2 + ( y - 1 )2 + 2018 ≥ 2018 ∀ x, y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
=> MinA = 2018 <=> x = -3 ; y = 1
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 \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
A = x^2 + 5y^2 + 4xy - 2y - 3
= x^2 + 4xy + 4y^2 + y^2 - 2y + 1 - 4
= ( x + 2y )^2 + ( y - 1 )^2 - 4 >= -4
Dấu ''='' xảy ra khi y = 1 ; x = -2
Vậy GTNN A là -4 khi x = -2 ; y = 1