Question

[Phạm Thị Phương Anh]

Cho x,y,z thoả mãn xy+yz+xz=1

Hãy tính giá trị biểu thức A= x\(\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}\)+ y\(\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}\)+ z \(\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)

Answer 1

[Võ Thị Quỳnh Giang]

ta có: xy+yz+zx=1

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

c/m tương tự ta đc: \(1+y^2=\left(x+y\right)\left(y+z\right)\)

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

thay vào A ta đc:

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

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

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

\(\Rightarrow A=2\) vì xy+yz+zx=1