Đáp án: $P=2$
Giải thích các bước giải:
Ta có:
$1+x^2=xy+yz+zx+x^2=y(x+z)+x(x+z)=(x+y)(x+z)$
Tương tự $1+y^2=(y+x)(y+z), 1+z^2=(z+x)(z+y)$
$\to P=x\sqrt{\dfrac{(y+x)(y+z)(z+x)(z+y)}{(x+y)(x+z)}}+y\sqrt{\dfrac{(z+x)(z+y)(x+y)(x+z)}{(y+x)(y+z)}}+z\sqrt{\dfrac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y))}}$
$\to P=x(y+z)+y(z+x)+z(x+y)$
$\to P=2(xy+yz+zx)$
$\to P=2$