x . y . x . z . y . z = 3 . 4 . 6

(x . x) . (y . y) . (z . z) = 72

x² . y² . z²  = 72

=>A=72

\(xy=3;xz=4;yz=6\Rightarrow xy.xz.yz=3.4.6\Leftrightarrow\left(xyz\right)^2=72\)\(\Leftrightarrow xyz=\pm6\sqrt{2}\)

+)\(xyz=-6\sqrt{2}\) => \(x=-\sqrt{2};y=-\frac{3\sqrt{2}}{2};z=-2\sqrt{2}\)

Thay vào A

+))\(xyz=6\sqrt{2}\) => \(x=\sqrt{2};y=\frac{3\sqrt{2}}{2};z=2\sqrt{2}\)

Thay vào A

\(\left\{\begin{matrix}xy=3\left(1\right)\\xz=4\left(2\right)\\yz=6\left(3\right)\end{matrix}\right.\).Từ \(yz=6\Rightarrow z=\frac{6}{y}\) thay vào (2) ta có:

\(xz=4\Rightarrow x\cdot\frac{6}{y}=4\)\(\Leftrightarrow\frac{6x}{y}=4\Leftrightarrow6x=4y\Leftrightarrow y=\frac{6x}{4}=\frac{3x}{2}\) thay vào (1) ta có:

\(x\cdot\frac{3x}{2}=3\Leftrightarrow\frac{3x^2}{2}=3\Leftrightarrow3x^2=6\Leftrightarrow x^2=2\)

Từ \(\left(1\right)\Rightarrow x^2y^2=9\Rightarrow y^2=\frac{9}{x^2}=\frac{9}{2}\)

Từ \(\left(2\right)\Rightarrow x^2z^2=16\Rightarrow z^2=\frac{16}{x^2}=\frac{16}{2}=8\)

Khi đó \(A=x^2+y^2+z^2=2+\frac{9}{2}+8=\frac{29}{2}\)

Ta có: \(x^2+y^2-z^2\)

\(=\left(x+y\right)^2-z^2-2xy\)

\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)

\(=-2xy\)

Ta có: \(x^2+z^2-y^2\)

\(=\left(x+z\right)^2-y^2-2xz\)

\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)

\(=-2xz\)

Ta có: \(y^2+z^2-x^2\)

\(=\left(y+z\right)^2-x^2-2yz\)

\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)

\(=-2yz\)

Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)

\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)

\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)

\(=\dfrac{-3}{2}\)

Ta có: \(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)

\(\left(y-z\right)\ge0\Leftrightarrow y^2+z^2\ge2yz\)

\(\left(z-x\right)^2\ge0\Leftrightarrow z^2+x^2\ge2zx\)

\(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\)

\(\left(y-1\right)^2\ge0\Leftrightarrow y^2+1\ge2y\)

\(\left(z-1\right)^2\ge0\Leftrightarrow z^2+1\ge2z\)

Cộng lại vế với vế ta được: 

\(3\left(x^2+y^2+z^2\right)+3\ge2xy+2yz+2zx+2x+2y+2z\)

\(\Leftrightarrow Q\ge\frac{2\left(x+y+yz+xy+yz+zx\right)-3}{3}=3\)

Dấu \(=\)khi \(x=y=z=1\).

Ta có: \(x+y+z+xy+yz+xz\le x+y+z+\frac{\left(x+y+z\right)^2}{3}\)

=> \(\left(x+y+z\right)^2+3\left(x+y+z\right)\ge3.6=18\)

<=> \(\left(x+y+z\right)^2+3\left(x+y+z\right)-18\ge0\)

<=> \(\left(x+y+z-3\right)\left(x+y+z+6\right)\ge0\)

<=> \(x+y+z\ge3\)(vì x + y + z + 6 > 0 vì x,y,z > 0)

Do đó: \(Q=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)

Dấu "=" xảy ra<=> x  = y= z và x + y + z = 3 <=> x = y = z = 1

Vậy MinQ = 3 <=> x = y= z = 1

Đề thiếu kìa :vv

⇔xy+yz+zx=0

=yz/(x−y)(x−z)

Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)