có: x+y+z=2=>(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=4

mà x^2+y^2+z^2=2 =>2(xy+yz+xz)=2

=>xy+yz+xz=1

xét:1+y^2=xy+yz+xz+y^2=(x+y)(z+y)

tương tự :1+z^2=xy+yz+xz+z^2=(x+z)(y+z)

1+x^2=xy+yz+xz+x^2=(x+z)(x+y)

thay vào M ta có :M=\(\sqrt{\frac{\left(x+y\right)\left(z+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}=\sqrt{\left(y+z\right)^2}\)=/y+z/

Mà x,y,z,\(\in\)Q=>đpcm

cho x,y >0 rút gọn A=\(\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}}}\)