Question

Cho ba số x,y,z thõa: xyz=1 tính:

\(M=\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)

Answer 1

cũng dễ thôi

M=\(\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\)

\(M=\dfrac{z}{z\left(1+x+xy\right)}+\dfrac{xz}{xz\left(1+y+yz\right)}+\dfrac{xyz}{xyz\left(1+z+zx\right)}\\ =\dfrac{z}{z+xz+xyz}+\dfrac{xz}{xz+xyz+xyz\left(z\right)}+\dfrac{xyz}{xyz+xyz\left(z\right)+xyz\left(xz\right)}\\ màxyz=1\\ nênM=\dfrac{z}{z+xz+1}+\dfrac{xz}{z+xz+1}+\dfrac{1}{z+xz+1}\\ vậyM=\dfrac{z+xz+1}{z+xz+1}=1\)