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\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)
Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)
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\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)
\(\left(x+4\right)^2\ge0\)
\(\left(x+4\right)^2-21\ge-21\)
\(-\left[\left(x+4\right)^2-21\right]\le21\)
Vậy Max B = 21 khi x = - 4
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\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)
\(\left(x-1\right)^2\ge0\)
\(\left(2y-1\right)^2\ge0\)
\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)
\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)
Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)
\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)
\(A_{min}=3\) khi \(x=-2\)
\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)
\(B_{min}=1\) khi \(x=10\)
\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)
\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)
\(A=x^2-10x+26\)
\(=\left(x^2-10x+25\right)+1\)
\(=\left(x-5\right)^2+1\ge1\)
Vậy \(Min_A=1\) khi \(x-5=0\Rightarrow x=5\)
\(B=x^2+7x+10=\left(x^2+7x+\dfrac{49}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{7}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)Vậy \(Min_B=\dfrac{-9}{4}\) khi \(x+\dfrac{7}{2}=0\Rightarrow x=\dfrac{-7}{2}\)
\(C=4x^2+8x+15=4\left(x^2+2x+1\right)+11=4\left(x+1\right)^2+11\ge11\)Vậy \(Min_C=11\) khi \(x+1=0\Rightarrow x=-1\)
\(D=3x^2-7x+20=3\left(x^2-\dfrac{7}{3}x+\dfrac{49}{36}\right)+\dfrac{191}{12}=3\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{12}\ge\dfrac{191}{12}\)Vậy \(Min_D=\dfrac{191}{12}\) khi \(x-\dfrac{7}{6}=0\Rightarrow x=\dfrac{7}{6}\)
\(E=x^2-4xy+5y^2-22y+8\)
\(=\left(x^2-4xy+4y^2\right)+\left(y^2-22y+121\right)-113\)\(=\left(x-2y\right)^2+\left(y-11\right)^2-113\ge-113\)
Vậy \(Min_E=-113\) khi \(\left[{}\begin{matrix}x-2y=0\\x-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}11-2y=0\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2y=11\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=11\end{matrix}\right.\)
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 \(B_{Min}=-11\Leftrightarrow x=4\)
\(a,A=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ A=\left(x-2y\right)^2+10\left(x-2y\right)+5+\left(y-1\right)^2+2\\ A=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
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/ \(4a^2+4a+2=\left(4a^2+4a+1\right)+1=\left(2a+1\right)^2+1\ge1\)
Suy ra BT đạt giá trị nhỏ nhất bằng 1 tại a = -1/2
b/ \(x^2-4xy+5y^2+10x-22y=\left(x-2y+5\right)^2+\left(y-1\right)^2-26\ge-26\)
Suy ra BT đạt giá trị nhỏ nhất bằng -26 khi y = 1 , x = -3
C = x2 - 4xy + 5y2 + 10x - 22y + 28
= (x2 - 4xy + 4y2) + (10x - 22y) + 25 + y2 + 3
= (x - 2y)2 + 10(x - 2y) + 25 + y2 + 3
= (x - 2y + 5)2 + y2 + 3 \(\ge\)3
Dấu " = " xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x-2y+5=0\\y=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=5\\y=0\end{cases}}\)
Vậy Min C = 3 \(\Leftrightarrow\)x = 5; y = 0
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 \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)