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\(A=x^2-2xy+6y^2-12x+2y+45\)
\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)
\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Do : \(\left(x-y-6\right)^2\text{≥}0\) ∀\(xy\) ; \(5\left(y-1\right)^2\text{≥}0\text{∀}y\)
⇒ \(\left(x-y-6\right)^2+5\left(y-1\right)^2\text{ ≥}0\)
⇔ \(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\text{≥}4\)
⇒ \(A_{Min}=4."="\text{⇔}x=7;y=1\)
\(A=x^2-2xy-12x+6y^2+2y+45\)
\(=x^2-2x\left(y+6\right)+\left(y+6\right)^2-\left(y+6\right)^2+6y^2+2y+45\)
\(=\left(x-\left(y+6\right)\right)^2-y^2-12y-36+6y^2+2y+45\)
\(=\left(x-y-6\right)^2+5y^2-10y+5+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Vậy \(A_{min}=4\)khi \(y=1\)và \(x=7\)
a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)
Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)
\(\Rightarrow\left(2x-3\right)^2+91\ge91\)
hay A \(\ge91\)
Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)
<=> 2x-3=0
<=> 2x=3
<=> \(x=\frac{3}{2}\)
Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)
b) \(B=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)
Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)
Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)
\(C=2x^2+2xy+y^2-2x+2y+2\)
\(C=x^2+2x\left(y-1\right)+\left(y-1\right)^2+x^2+1\)
\(\Leftrightarrow C=\left(x+y-1\right)^2+x^2+1\)
Ta có:
\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)
\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)
hay C\(\ge\)1
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x+y-1\right)^2=0\\x^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=1\\x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=0\end{cases}}}\)
Vậy Min C=1 đạt được khi y=1 và x=0
A=\(\left(x-y\right)^2-2.6.\left(x-y\right)+36+5y^2+10y+5+4\)
=\(\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
Dấu bằng xảy ra khi y=1 và x=5
2B=\(2x^2+2y^2-2xy-2x+2y+2\)
=\(\left(x-y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
=>B\(\ge\)0
\(A=x^2-2xy+6y^2-12x+2y+54\)
\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)
\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)
Do: \(\left(x-y-6\right)^2\ge0\forall xy\); \(5\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-y-6\right)^2+5\left(y-1\right)^2\ge0\)
\(\Leftrightarrow A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
\(\Rightarrow A_{Min}=4\)
Dấu "=" xảy ra khi \(x=7;y=1\)
\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)
\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)\(\ge4\)
Amin=4 khi y=1; x=7
\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)
\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)
\(Amin=4\)\(khi\)\(y=1;x=7\)