Ta có: $(x+y+z)^2=25^2=625$
$x^2+y^2+z^2=233$
Suy ra $xy+yz+zx=196$
$x\sqrt{\dfrac{(196+y^2)(196+z^2)}{196+x^2}}= x \sqrt{\dfrac{(xy+yz+xz+y^2)(xy+yz+xz+z^2)}{xy+yz+xz+x^2}}=x\sqrt{\dfrac{(x+z)(x+y)(y+z)^2}{(x+z)(x+y)}}=x(y+z)$
Tương tự với mấy cái kia, ta được: $x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2.196=392$