Cho x,y,z là các số dương thỏa mãn : xy+yz+xz=2011 Tính : A= x$\sqrt{\frac{(2011+y^2)(2011+z^2)}{2011+x^2}}$ +y$\sqrt{\frac{(2011+x^2)(2011+z^2)}{2011+y^2}}$+z$\sqrt{\frac{(2011+y^2)(2011+x^2)}{2011+z^2}}$

Đáp án:

 `A=4022`

Giải thích các bước giải:

 `A=xsqrt{((2011+y^2)(2011+z^2))/(2011+x^2)}+ysqrt{((2011+x^2)(2011+z^2))/(2011+y^2)}+zsqrt{((2011+y^2)(2011+x^2))/(2011+z^2)}`

`+)2011+y^2`

`=y^2+xy+yz+zx`

`=y(x+y)+z(x+y)`

`=(x+y)(y+z)`

`CMT^2` ta cũng có

`2011+x^2`

`=(x+y)(x+z)`

`2011+z^2`

`=(y+z)(z+x)`

`=> A=xsqrt{((2011+y^2)(2011+z^2))/(2011+x^2)}+ysqrt{((2011+x^2)(2011+z^2))/(2011+y^2)}+zsqrt{((2011+y^2)(2011+x^2))/(2011+z^2)}`

`=xsqrt{((x+y)(y+z)(y+z)(z+x))/((x+y)(x+z))}+ysqrt{((x+y)(x+z)(y+z)(z+x))/((x+y)(y+z))}+zsqrt{((x+y)(y+z)(x+y)(z+x))/((y+z)(x+z))}`

`=xsqrt{(y+z)^2}+ysqrt{(x+z)^2}+zsqrt{(x+y)^2}`

`=x(y+z)+y(x+z)+z(y+x)`

`=xy+xz+xy+yz+zy+xz`

`=2(xy+yz+zx)`

`=2.2011`

`=4022`

Vậy `A=4022`

$@Kate2007$