Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(A\left(3,4\right),B\left(x,y\right),N\left(0,y\right),M\left(x,0\right)\).
Khi đó \(f\left(x,y\right)=\sqrt{\left(x-3\right)^2+\left(y-4\right)^2}+\left|x\right|+\left|y\right|\)
\(=BA+BM+BN\)
\(\ge BA+BO\)
\(\ge AO\)(theo bđt tam giác)
Dấu \(=\)khi \(B\equiv O\)suy ra \(x=y=0\).
Vậy \(minf\left(x,y\right)=f\left(0,0\right)=5\).
a/ \(f\left(x\right)=2x^2-2\left(a+b\right)x+a^2+b^2\)
\(=\frac{1}{2}\left[4x^2-4\left(a+b\right)x+\left(a+b\right)^2\right]+\frac{1}{2}a^2+\frac{1}{2}b^2-ab\)
\(=\frac{1}{2}\left(2x-a-b\right)^2+\frac{1}{2}\left(a-b\right)^2\ge\frac{1}{2}\left(a-b\right)^2\)
Dấu "=" xảy ra khi \(x=\frac{a+b}{2}\)
b/ \(f\left(x\right)=3x^2-2\left(a+b+c\right)x+a^2+b^2+c^2\)
\(=\frac{1}{3}\left[9x^2-6\left(a+b+c\right)x+\left(a+b+c\right)^2\right]+\frac{2}{3}\left(a^2+b^2+c^2\right)-\frac{2}{3}\left(ab+bc+ca\right)\)
\(=\frac{1}{3}\left(3a-a-b-c\right)^2+\frac{2}{3}\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow f\left(x\right)_{min}=\frac{2}{3}\left(a^2+b^2+c^2-ab-bc-ca\right)\) khi \(x=\frac{a+b+c}{3}\)
a/ \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y-5\right)^2\ge0\\\left(x-y+4\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow\left(x-1\right)^2+\left(y-5\right)^2+\left(x-y+4\right)^2\ge0\)
\(A_{min}=0\) khi \(\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
b/ \(B=x^2y^2-6xy+9+x^2+4x+4-16\)
\(B=\left(xy-3\right)^2+\left(x+2\right)^2-16\ge-16\)
\(B_{min}=-16\) khi \(\left\{{}\begin{matrix}x=-2\\y=-\frac{3}{2}\end{matrix}\right.\)
c/ \(C=x^2+\frac{y^2}{4}+16+xy+8x+4y+\frac{59}{4}y^2-3y+2001\)
\(C=\left(x+\frac{y}{2}+4\right)^2+\frac{59}{4}\left(y-\frac{6}{59}\right)^2+\frac{118050}{59}\ge\frac{118050}{59}\)
\(C_{min}=\frac{118050}{59}\)
d/ \(D=\left(x^2-2x\right)\left(y^2+6y\right)+12\left(x^2-2x\right)+3\left(y^2+6y\right)+36\)
\(=\left(x^2-2x\right)\left(y^2+6y+12\right)+3\left(y^2+6y+12\right)\)
\(=\left(x^2-2x+3\right)\left(y^2+6y+12\right)\)
\(=\left[\left(x-1\right)^2+2\right]\left[\left(y+3\right)^2+3\right]\ge2.3=6\)
\(D_{min}=6\)
e/ \(E=a^2+\frac{b^2}{4}+\frac{9}{4}+ab-3a-\frac{3b}{2}+\frac{3b^2}{4}-\frac{3b}{2}+2014-\frac{9}{4}\)
\(=\left(a+\frac{b}{2}-\frac{3}{2}\right)^2+\frac{3}{4}\left(y-1\right)^2+2011\ge2011\)
\(E_{min}=2011\)
\(A=\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|\)
\(=\left|x-2012\right|+\left|2014-x\right|+\left|x-2010\right|\)
\(\ge\left|x-2012\right|+\left|2014-x+x-2010\right|\)
\(=\left|x-2012\right|+4\)
Vì \(\left|x-2012\right|\ge0\forall x\)
\(\Rightarrow\left|x-2012\right|+4\ge4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=2012\)
Vậy MIN \(A=4\Leftrightarrow x=2012\)