Question

[Bạch Liên]

Cho x, y, z > 0 và xy + yz + zx = 1

Tính giá trị biểu thức :

A = x.  √ [(1+y²)(1+z²)] / √(1+x²) + y .  √[(1+z²)(1+x²)] / √(1+y²) + z.  √[(1+x²)(1+y²)] / √(1+z²)

Answer 1

[ミ★kͥ-yͣeͫt★彡]

Ta có: \(1+x^2=xy+yz+xz+x^2=\left(x+y\right)\left(x+z\right)\)

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

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

Thay vào biểu thức A, ta có bt sau:

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

\(+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)

\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(z+y\right)}}\)

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

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)(x,y,z dương)

\(=2\left(xy+xz+yz\right)=2.1=2\)