Question

Cho x,y,z thỏa mãn xy+yz+xz = 1.Tính

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

Answer 1

Lời giải:

Ta thấy: \(xy+yz+xz=1\)

\(\Rightarrow \left\{\begin{matrix} 1+y^2=xy+yz+xz+y^2=(y+z)(y+x)\\ 1+x^2=xy+yz+xz+x^2=(x+y)(x+z)\\ 1+z^2=xy+yz+xz+z^2=(z+x)(z+y)\end{matrix}\right.\)

Do đó:

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

Hoàn toàn tt:

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

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

Cộng theo vế:

\(S=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)

Answer 2

Lời giải:

Ta thấy: xy+yz+xz=1

⇒⎧⎪⎨⎪⎩1+y2=xy+yz+xz+y2=(y+z)(y+x)1+x2=xy+yz+xz+x2=(x+y)(x+z)1+z2=xy+yz+xz+z2=(z+x)(z+y)

Do đó:

x√(y2+1)(z2+1)1+x2=x√(y+x)(y+z)(z+x)(z+y)(x+y)(x+z)=x√(y+z)2=x(y+z)

Hoàn toàn tt:

y√(x2+1)(z2+1)y2+1=y(x+z) 

z√(x2+1)(y2+1)z2+1=z(x+y)

Cộng theo vế:

S=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2