Question

[Trà My]

Cho ba số dương x,y,z  thõa mãn điều kiện xy+yz+zx=1 tính:

\(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)

Answer 1

[Vũ Quỳnh Anh]

bài lớp mấy mà khó dữ

Answer 2

[ღᏠᎮღ🆃🆄ấ🅽ঔ 🅽🅰🅼ঌ❄๖ۣۜ]

Ta có : \(xy+yz+zx=1\)

\(\Rightarrow\hept{\begin{cases}1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\\1+y^2=xy+yz+zx+y^2=\left(y+x\right)\left(y+z\right)\\1+z^2=xy+yz+zx+z^2=\left(z+x\right)\left(z+y\right)\end{cases}}\)

Do đó :

\(\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=\sqrt{\left(y+z\right)^2}\)\(=y+z\)

\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\left(y+z\right)\)

Hoàn toàn tương tự :

\(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\left(z+x\right)\)

\(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\left(x+y\right)\)

Do đó :

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\)

\(=2\left(xy+yz+zx\right)=2\)