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

xy=4 , xz=6

=> \(x^2zy\)= 3x4=12

=>\(x^2\)=\(12:yz=12:6=2\)

\(xz=4,yz=6\)

=>\(z^2xy=6x4=24\)

=>\(z^2=24:xy=8\)

\(xy=3,yz=6\)

=>\(xy^2z=6x3=18\)

=>\(y^2=18:xz=18:4=4.5\)

Vậy \(x^2+y^2+z^2=2+8+4.5=14.5\)

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

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

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

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

\(P=11\)

Đề thiếu kìa :vv

⇔xy+yz+zx=0

=yz/(x−y)(x−z)

Tương tự: xy/z^2+2xy=xy/(x−z)(y−z)

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

Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)

Cộng VTV:

\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)