1) \(A=x^2-4x+1\)

\(A=x^2-4x+4-3\)

\(A=\left(x^2-4x+4\right)-3\)

\(A=\left(x-2\right)^2-3\)

Ta có: \(\left(x-2\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-2\right)^2-3\ge-3\) với mọi x

Vậy MIinA = -3 khi x = 2

2) \(B=-x^2+13x+2012\)

\(B=-x^2+13x-\frac{169}{4}+\frac{169}{4}+2012\)

\(B=-\left(x^2-13+\frac{169}{4}\right)+\left(\frac{169}{4}+2012\right)\)

\(B=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\)

Ta có: \(\left(x-\frac{13}{2}\right)^2\ge0\) với mọi x

\(-\left(x-\frac{13}{2}\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\)

Vây \(Max\left(B\right)=\frac{8217}{4}\) khi \(x=\frac{13}{2}\)

1/ Ta có : P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}P(x)=−x2+13x+2012=−(x−213​)2+48217​≤48217​
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2

1/ Ta có : P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}P(x)=−x2+13x+2012=−(x−213​)2+48217​≤48217​
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
2/ Ta có : x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1x3+3xy+y3=x3+3xy.1+y3=x3+y3+3xy(x+y)=(x+y)3=1
3/ a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0a+b+c=0⇔(a+b+c)2=0⇔a2+b2+c2+2(ab+bc+ac)=0
\Leftrightarrow ab+bc+ac=-\frac{1}{2}⇔ab+bc+ac=−21​ \Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}⇔(ab+bc+ac)2=41​⇔a2b2+b2c2+c2a2+2abc(a+b+c)=41​
\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}⇔a2b2+b2c2+c2a2=41​(vì a+b+c=0)
Ta có : a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1a2+b2+c2=1⇔(a2+b2+c2)2=1⇔a4+b4+c4+2(a2b2+b2c2+c2a2)=1
\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}⇔a4+b4+c4=1−2(a2b2+b2c2+c2a2)=1−42.1​=21​

\(P\left(x\right)=-x^2+13x-42,25+1969,75\)

\(P\left(x\right)=-\left(x^2-2\cdot6.5\cdot x+6.5^2\right)+1969,75\)

\(P\left(x\right)=-\left(x-6,5\right)^2+1969,75\le1969,75\)

Dấu \("="\) xảy ra khi \(x-6,5=0\Rightarrow x=6,5\)

Vậy Max_P=1969,75 khi x=6,5

\(C=13x+2012-x^2\)

\(=-\left(x^2-13x+\dfrac{169}{4}\right)+\dfrac{7879}{4}\)

\(=-\left(x-\dfrac{13}{2}\right)^2+\dfrac{7879}{4}\)

Nhận xét :

\(\left(x-\dfrac{13}{2}\right)^2\ge0\)

\(\Leftrightarrow-\left(x-\dfrac{13}{2}\right)^2\le0\)

\(\Leftrightarrow-\left(x-\dfrac{13}{2}\right)+\dfrac{7879}{4}\le\dfrac{7879}{4}\)

\(\Leftrightarrow C\le\dfrac{7879}{4}\)

Dấu "=" xảy ra khi : \(\left(x-\dfrac{13}{2}\right)^2=0\Leftrightarrow x=\dfrac{13}{2}\)

Vậy...

A = -x^2 + 13x + 2012

4A = -4x^2 + 52x + 8048

4A = -(2x-13)^2 + 8048 + 169

4A = -(2x-13)^2 + 8217

A = -(2x-13)^2/4 + 2054,25

GTLL của A = 2054,25 khi 2x-13 = 0 <=> x=6,5

1/ Ta có : \(P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\)

Dấu "=" xảy ra khi x = 13/2

Vậy Max P(x) = 8217/4 tại x = 13/2

2/ Ta có : \(x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1\)

3/ \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow ab+bc+ac=-\frac{1}{2}\) \(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)(vì a+b+c=0)

Ta có : \(a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)

\(\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}\)

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