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13:
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\\ \Leftrightarrow\left(x+y+z\right)\left(xy+yz+zx\right)-xyz=0\\ \Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
\(\forall x=-y\Leftrightarrow VT=-y^{2017}+y^{2017}+z^{2017}=z^{2017}=\left(-y+y+z\right)^{2017}=VP\\ \forall y=-z\Leftrightarrow VT=x^{2017}-z^{2017}+z^{2017}=x^{2017}=\left(x-z+z\right)^{2017}=VP\\ \forall z=-x\Leftrightarrow VT=x^{2017}+y^{2017}-x^{2017}=y^{2017}=\left(x+y-x\right)^{2017}=VP\)
Vậy ta đc đpcm
ko vt lại đề
(xyz-xy)-(yz-y)-(zx-x)+(z-1)=2019
=>xy(z-1)-y(z-1)-x(z-1)+(z-1)=2019
=> (z-1)(xy-y-x+1)=2019
=> (z-1)(z-1)(y-1)=2019
vì x>y>z>0 => (x-1) khác (y-1) khác (z-1)=> x-1>y-1>z-1
nên (z-1),(x-1)và (y-1) thuộc ước của 2019={ 1,3,673,2019}
(x-1)(y-1)(z-1)= 673.3.1=2019
=> x-1=673=>x=674
=>y-1=3=>y=4
=> z-1 =1=>z=2
Vậy x=674,y=4,z=2