Question

Cho  xy +yz +xz = 1 

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

Answer 1

1 + x² = xy + yz + zx + x² = y(x+z) + x(z+x) = (x+y).(x+z)

Tương tự, 1 + y² = (y + x). (y +z) và 1 + z² = (z +x).(z+y)

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

Tương tự => A = x |y +z| + y.|x+ z| + z.|x+y| 

Có thể đề là rút gọn A. Yêu cầu tính A, không đủ dữ kiện ( Vid dụ : Nếu y + z > 0 và x + z< 0; x+ y < 0 => A = -2yz)

Nếu Thêm điều kiện x; y; z > 0 => A = x(y+z) + y(x+z) + z(x+y) = 2(xy + yz+ zx) = 2

 

Answer 2

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

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

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

\(\text{Suy ra: }A=x\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+y\right)\left(x+z\right)\left(z+x\right)\left(z+y\right)}{\left(y+x\right)\left(y+z\right)}}\)

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

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

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