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\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2015}\)
\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\) (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
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{-x-y}{\left(x+y+z\right)z}\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{\left(x+y+z\right)z}\right)=0\)
\(+,x+y=0\Rightarrow x=-y\Rightarrow\text{đpcm}\)
\(+,\frac{1}{xy}+\frac{1}{\left(x+y+z\right)z}=0\Leftrightarrow\frac{xy+xz+yz+z^2}{xyz\left(x+y+z\right)}=0\Leftrightarrow\frac{x\left(y+z\right)+z\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\frac{\left(y+z\right)^2}{xyz\left(x+y+z\right)}=0\Rightarrow y+z=0\Rightarrow z=-y\Rightarrow\text{đpcm}\)
\(\text{Vậy ta có điều phải chứng minh }\)
Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\) \(\frac{yz+xz+xy}{xyz}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\) \(yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)=xyz\)
\(\Leftrightarrow\) \(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)
\(\Leftrightarrow\) \(2xyz+y^2z+yz^2+x^2z+xz^2+x^2y+xy^2=0\)
\(\Leftrightarrow\) \(x^2\left(y+z\right)+x\left(y^2+2yz+z^2\right)+yz\left(y+z\right)=0\)
\(\Leftrightarrow\) \(\left(y+z\right)\left(x^2+xy+xz+yz\right)=0\)
\(\Leftrightarrow\) \(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\) \(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
Vậy, trong ba số x, y, z có hai số đối nhau
\(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y=2xyz\)\(\Leftrightarrow\left(x^2y-xy^2\right)+\left(x^2z-xyz\right)+\left(z^2y-z^2x\right)+\left(y^2z-xyz\right)=0\)\(\Leftrightarrow xy\left(x-y\right)+xz\left(x-y\right)-z^2\left(x-y\right)-yz\left(x-y\right)=0\)\(\Leftrightarrow\left(xy+xz-z^2-yz\right)\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[x\left(y+z\right)-z\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-z\right)\left(y+z\right)=0\Rightarrow\left[{}\begin{matrix}x-y=0\\x-z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\Rightarrowđpcm\)
\(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y=2xyz\)
\(x^2y-y^2x+x^2z-z^2x+y^2z+z^2y-2xyz=0\)
\(\left(x^2y-y^2x\right)+\left(x^2z-xyz\right)+\left(z^2y-z^2x\right)=\left(y^2z-xyz\right)+\left(y^2z-xyz\right)=0\)
\(\left[\left(x-y\right)\left(xy\right)\right]+\left[\left(x-y\right)\left(zx\right)\right]+\left[\left(x-y\right)\left(-z^2\right)\right]+\left[\left(x-y\right)\left(-yz\right)\right]\)
\(\left(x-y\right)\left(xy+zx-z^2-yz\right)=\left(x-y\right)\left(x-z\right)\left(y+z\right)\)
đpcm
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow x^2y+y^2z+x^2z+y^2x+z^2y+z^2x+2xyz=0\)
\(\Leftrightarrow\left(x^2y+y^2x\right)+\left(x^2z+xyz\right)+\left(z^2x+z^2y\right)+\left(y^2z+xyz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\left(xy+yz\right)+\left(xz+z^2\right)\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
Vậy 2 trong 3 số x,y,z có 2 số đối nhau