a)\(P=-x^2-4x+16\)

\(=-x^2-4x-4-12\)

\(=-\left(x^2+4x+4\right)-12\)

\(=-\left(x+2\right)^2-12\le-12\)

Đẳng thức xảy ra khi \(x=-2\)

b)\(-x^2+2xy-4y^2+2x+10y-2017\)

\(=\left(-x^2+2xy-y^2+2x-2y-1\right)+\left(-3y^2+12y-12\right)-2004\)

\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)-2004\)

\(=-\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]-3\left(y-2\right)^2-2004\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2-2004\le-2004\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Nỗi hứng lm cho vui!

Bài 1:

a) H = \(x^2-4x+16=\left(x^2-4x+4\right)+12=\left(x-2\right)^2+12\)

Vì \(\left(x-2\right)^2\ge0\) => H \(\ge\) 12

=> Dấu = xảy ra <=> \(x=2\)

b) K = \(2x^2+9y^2-6xy-8x-12y+2018\)

= \(\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+\left(x^2-12x+36\right)+1982\)

= \(\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-6\right)^2+1978\)

= \(\left(x-3y+2\right)^2+\left(x-2\right)^2+1978\)

Vì \(\left\{{}\begin{matrix}\left(x-3y+2\right)^2\ge0\\\left(x-6\right)^2\ge0\end{matrix}\right.\) => K \(\ge\) 1978

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)

Bài 2:

a) P = \(-x^2-4x+16=-\left(x^2+4x+4\right)+20\)

= \(-\left(x+2\right)^2+20\le20\)

=> Dấu = xảy ra <=> \(x=-2\)

b) \(Q=-x^2+2xy-4y^2+2x+10y-2017\)

= \(-\left[\left(x^2-2xy+y^2\right)+3\left(y^2-4y+4\right)-2\left(x-y\right)+2005\right]\)

= \(-\left[\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2004\right]\)

= \(-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2\right]-2004\)

Vì \(\left\{{}\begin{matrix}-\left(x-y-1\right)^2\le0\\3\left(y-2\right)^2\le0\end{matrix}\right.\) => Q \(\le-2004\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x=y+1\\y=2\end{matrix}\right.\) <=> \(x=3;y=2\)

đăng từng câu 1 thôi

câu A thiếu đề

B=\(x^2-2x+2017=\left(x-1\right)^2+2016>=2016\)

Min B=2016 khi x-1=0<=>x=1

+)D=\(-2x^2+4x+2017=-2\left(x^2-2x+1\right)+2019=-2\left(x-1\right)^2+2019< =2019\)

=>Max D=2019, dấu '=' xảy ra khi x-1=0<=>x=1

Bổ sung câu A. \(A=x^2+2xy+3y^2-4y+2017\)

\(A=x^2+2xy+3y^2-4y+2017\)

\(A=\left(x^2+2xy+y^2\right)+\left(2y^2-4y+2\right)+2015\)

\(A=\left(x^2+2xy+y^2\right)+2\left(y^2-2y+1\right)+2015\)

\(A=\left(x+y\right)^2+2\left(y-1\right)^2+2015\ge2015\)

Vậy A_min=2015 <=> x=-1 và y=1

\(A=-\left(x^2-2x\left(y+1\right)+\left(y+1\right)^2\right)-\left(4y^2-10y-5-\left(y+1\right)^2\right)\)

\(=-\left(x-y-1\right)^2-\left(3y^2-12y-6\right)\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+18\le18\)

Max A=18 khi y=2; x=3

\(B=-\left(x^2+2x\left(y-1\right)+\left(y-1\right)^2\right)-\left(2y^2+2y-\left(y-1\right)^2\right)-15\)

\(=-\left(x+y-1\right)^2-\left(y+2\right)^2-10\le-10\)

Max B=-10 khi y=-2; x= 3

a ) \(x^2-x+1\)

\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)

Bài 1 :

=-5(x^2+4/5x+19/25)

=-5(x^2+2x.2/5+4/25+3/5)

=-5(x+2/5)^2-3

Vì (x+2/5)^2 lớn hơn hoặc bằng 0 =>-5(x+2/5)^2-3 nhỏ hơn hoặc bằng-3

Vậy Min là-3