a)\(P=x^2+4x+2xy+3y^2+5y+2017\)

\(=x^2+2xy+y^2+4y+4+4x+2y^2+y+\dfrac{1}{8}+\dfrac{16103}{8}\)

\(=\left(x+y+2\right)^2+2\left(y^2+\dfrac{y}{2}+\dfrac{1}{16}\right)+\dfrac{16103}{8}\)

\(=\left(x+y+2\right)^2+2\left(y+\dfrac{1}{4}\right)^2+\dfrac{16103}{8}\ge\dfrac{16103}{8}\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{7}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)

b)\(Q=-x^2+4x-3y^2+6y+2017\)

\(=-x^2+4x-4-3y^2+6y+3+2024\)

\(=-\left(x^2-4x+4\right)-\left(3y^2-6y-3\right)+2024\)

\(=-\left(x-2\right)^2-3\left(y^2-2y-1\right)+2024\)

\(=-\left(x-2\right)^2-3\left(y-1\right)^2+2024\ge2024\)

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

Ta có:

\(P=x^2+4x+2xy+3y^2+5y+2017\)

\(=x^2+2x\left(y+2\right)+\left(y+2\right)^2+2y^2+y+2013\)

\(=\left[x+\left(y+2\right)\right]^2+2\left(y^2+y+0,25\right)+2012,5\)

\(=\left(x+y+2\right)^2+2\left(y+0,5\right)^2+2012,5\ge2012,5\)

Dấu "=" xảy ra khi:

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2=0\\y+0,5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-0,5\\x=-1,5\end{matrix}\right.\)

Vậy \(minP=2012,5\) khi \(\left\{{}\begin{matrix}y=-0,5\\x=-1,5\end{matrix}\right.\)

Ta có:

\(Q=-x^2+4x-3y^2+6y+2017\)

\(=-\left(x^2-4x+4\right)-3\left(y^2-2y+1\right)+2024\)

\(=-\left(x-2\right)^2-3\left(y-1\right)^2+2024\le2024\)

Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(maxQ=2024\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

bạn xem trong danh sách câu trả lời của mình ấy, mình đã trả lời nhiều bài tương tự rồi

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

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

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

\(A=3x^2+5x-2\)

\(A=3\left(x^2+\frac{5}{3}x-\frac{2}{3}\right)\)

\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2-\frac{49}{36}\right)\)

\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2\right)-\frac{49}{12}\)

\(A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\)

         Vì \(3\left(x+\frac{5}{6}\right)^2\ge0\)

                  Do đó \(3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)

Dấu = xảy ra khi \(x+\frac{5}{6}=0\Rightarrow x=-\frac{5}{6}\)

      Vậy Min A=\(-\frac{49}{12}\) khi x=\(-\frac{5}{6}\)

mk làm ý a thôi, mấy ý sau dựa vào mà làm.

      A = \(3x^2+5x-2\)

 => \(\frac{A}{3}=x^2+\frac{5}{3}x-\frac{2}{3}\)(chia cả 2 vế cho 3)

\(\Leftrightarrow\frac{A}{3}=x^2+2.x.\frac{5}{6}+\left(\frac{5}{6}\right)^2-\frac{49}{36}\)

\(\Leftrightarrow\frac{A}{3}=\left(x+\frac{5}{6}\right)^2-\frac{49}{36}\)

\(\Rightarrow A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\ge-\frac{49}{12}\)

Đẳng thức xảy ra <=> x = - 5/6.

Vậy Min A = - 49/12 khi và chỉ khi x = - 5/6.

mk ko biết, nhìn hoi phức tạp nhỉ