Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+3xyz-xyz=0\)

\(\Leftrightarrow\left(x^2y+xy^2\right)+\left(yz^2+z^2x\right)+\left(zx^2+2xyz+y^2z\right)=0\)

\(\Leftrightarrow xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(xy+z^2+yz+zx\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=> x = -y hoặc y = -z hoặc z = -x

Không mất tổng quát giả sử x = -y, khi đó:

\(\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=-\frac{1}{y^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{z^{2015}}\)

\(\frac{1}{x^{2015}+y^{2015}+z^{2015}}=\frac{1}{-y^{2015}+y^{2015}+z^{2015}}=\frac{1}{z^{2015}}\)

\(\Rightarrow\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{x^{2015}+y^{2015}+z^{2015}}\)

Từ x+y+z=2015 => \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z\Rightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Rightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\Rightarrow\frac{\left(x+y\right)\left(xy+yz+zx+z^2\right)}{xyz\left(x+y+z\right)}=0\)

\(\Rightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)(Do x,y,z khác 0)

Mà x+y+z=2015 và (x+y)(y+z)(x+z)=0

=> x+y=0 => z =2015

hoặc y+z=0 => x=2015

hoặc x+z=0 => y=2015

                         Vậy nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z=2015\)thì ít nhất 1 trong 3 số x,y,z bằng 2015(ĐPCM)

               lik.e nhé!

đề có sai k vậy bạn?

Từ giả thiết ta có ngay \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Suy ra x + y = 0 hoặc y + z = 0 hoặc z + x = 0

Tới đây bạn tự làm nhé :)

        \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2015}\)

\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)   (do x+y+z = 2015)

\(\Rightarrow\)\(\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)

\(\Rightarrow\)\(\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)

\(\Rightarrow\)\(\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\)

\(\Rightarrow\)\(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

đến đây tự lm nốt nha