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

Do x; y ; z > 0 nên xyz khác 0 => \(\frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=1\Rightarrow\frac{1}{z}+\frac{1}{x}+\frac{1}{y}=1\Rightarrow\frac{1}{x}1\)

Vì x<= y< = z nên \(\frac{1}{x}\ge\frac{1}{y}\ge\frac{1}{z}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}\)

=> 1 < = 3/x => x < = 3 mà x > 1 nên x = 2 hoặc 3

Nếu x = 2 => \(\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\Rightarrow\frac{1}{y}2;\frac{1}{y}+\frac{1}{z}\le\frac{2}{y}\Rightarrow\frac{2}{y}\ge\frac{1}{2}\Rightarrow y\le4\)

mà y >2 => y = 3 hoặc 4 

y = 3 => z = 6;

y = 4 => z = 4

nếu x = 3 => \(\frac{1}{y}+\frac{1}{z}=\frac{2}{3}\Rightarrow\frac{1}{y}\frac{3}{2};\frac{1}{y}+\frac{1}{z}\le\frac{2}{y}\Rightarrow\frac{2}{y}\ge\frac{2}{3}\Rightarrow y\le3\)

theo đề bài x<= y nên y = 3 => z = 3

Vậy (x;y;z) = (3;3;3); (2;3;6);(2;4;4)

x=1;y=9;z=8

Kiểm tra lại mà xem.

x=2;y=3;z=6

x=3;y=3;z=3

bài này học còn gì hả hói

*Xét 0<x<y<z

Ta thấy: xy<yz (x<z)

             zx<yz (x<y)

=>xy+yz+zx=xyz<zy+zy+zy

=>xyz<3zy

=>x<3 mà 0<x<3

=>x=1;2

-Nếu x=1

=>y+yz+z=yz

=>y+z     =yz-yz

=>y+z      =0

mà 0<y<z

=>Vô lí

-Nếu x=2

=>2y+yz+2z=2yz

=>2y+2z      =2yz-yz

=>2.(y+z)    =yz

Ta thấy: y<z

=>2.(y+z)=yz<2.(z+z)

=>yz<4z

=>y<4 mà 2<y<4

=>y=3

=>2.3+3z+2z=2.3.z

=>6+5z         =6z

=>z               =6

*Xét0<x=y=z

=>xx+xx+xx=xxx

=>3xx          =xxx

=>x              =3

=>x=y=z=3

Vậy x=2;y=3;z=6

       x=3;y=3;z=3

x=2;y=3;z=6

x=3;y=3;z=3

\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)

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)