Tìm max, min:
\(A=x^2+2xy-4y+2017\)
\(B=x^2-2x+2017\)
\(C=-4x^2+8xy-3y^2+y-2017\)
\(D=-2x^2+4x+2017\)
K
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TD
0
DT
2
23 tháng 6 2017
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 \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
23 tháng 6 2017
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 \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
DT
1
20 tháng 6 2017
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 \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
MD
16 tháng 6 2017
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 <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)
MD
16 tháng 6 2017
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 <=> \(\left\{{}\begin{matrix}x=y+1\\y=2\end{matrix}\right.\) <=> \(x=3;y=2\)
HT
1
17 tháng 6 2017
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 \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
NH
2
23 tháng 5 2017
\(A=3x^2+5x-2\)
\(A=3\left(x^2+\frac{5}{3}x-\frac{2}{3}\right)\)
\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2-\frac{49}{36}\right)\)
\(A=3\left(x^2+2.\frac{5}{6}x+\left(\frac{5}{6}\right)^2\right)-\frac{49}{12}\)
\(A=3\left(x+\frac{5}{6}\right)^2-\frac{49}{12}\)
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=\(-\frac{5}{6}\)
KH
23 tháng 5 2017
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.
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. \(A=x^2+2xy+3y^2-4y+2017\)