THAY 2018 = xyz vào biểu thức 

      \(\frac{xyzx}{xy+xyzx+xyz}\)  +  \(\frac{y}{yz+y+xyz}\)+  \(\frac{z}{xz+z+1}\)

 =  \(\frac{xz}{1+xz+z}\)+  \(\frac{1}{z+1+xz}\)+  \(\frac{z}{xz+z+1}\)=  \(\frac{xz+z+1}{xz+z+1}\)=\(1\)

Đặt \(A=\frac{2018x}{xy+2018x+2018}+\frac{y}{yzz+y+2018}+\frac{z}{xz+z+1}\)

Thay \(xyz=2018\)vào A ta được 

\(A=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

   \(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{1}{xz+z+1}\)

  \(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

  \(=\frac{xz+1+z}{xz+z+1}=1\)

thay xyz=2018 vào M ta có

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

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+x+1}\)

\(=\frac{x^2yz}{xy\left(1+xz+y\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+x+1}\)

\(=\frac{xz}{1+xz+y}+\frac{1}{z+1+xz}+\frac{z}{xz+1+xz}=\frac{xz+1+z}{z+1+xz}=1\)

Vậy M=1 với xyz=2018

Em chỉ làm đại thôi ạ, có gì sai mong chị bảo vì năm nay em mới lên lớp 7 :vv

\(M=\frac{2018x}{xy+2018x+2018}+\frac{y}{yz+y+2018}+\frac{z}{xz+z+1}\)

\(=\frac{2018x}{xy+2018x+2018}+\frac{xy}{xyz+xy+2018x}+\frac{xyz}{xyxz+xyz+xy}\)

\(=\frac{2018x}{xy+2018x+2018}+\frac{xy}{2018+xy+2018x}+\frac{2018}{xy+2018+2018x}\)

\(=\frac{2018x+xy+2018}{xy+2018x+2018}=1\)

Vậy M = 1.

Thế xyz = 2018 vào M ta được :

\(M=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(M=\frac{x^2yz}{xy\left(1+zx+z\right)}+\frac{y}{y\left(1+zx+z\right)}+\frac{z}{zx+z+1}\)

\(M=\frac{x^2y^2z}{xy^2\left(1+zx+z\right)}+\frac{y^2x}{xy^2\left(1+zx+z\right)}+\frac{xy^2z}{xy^2\left(1+zx+z\right)}\)

\(M=\frac{x^2y^2z+y^2x+xy^2z}{xy^2\left(1+zx+z\right)}\)

\(M=\frac{xy^2\left(zx+1+z\right)}{xy^2\left(1+zx+z\right)}\)

\(M=1\)

Vậy M = 1 khi xyz = 2018

Đặt biểu thức trên là A, thay xyz = 2018, ta dược :

\(A=\dfrac{x^2yz}{xy+xyz+x^2yz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+x+1}\)

\(=\dfrac{xy\left(xz\right)}{xy\left(1+z+xz\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{z+zx+1}\)

\(=\dfrac{xz}{1+z+xz}+\dfrac{1}{z+1+xz}+\dfrac{z}{z+zx+1}=\dfrac{xz+1+z}{1+z+xz}=1\)

⇒ĐPCM

Please help me!!!!!!!!!!!

I feel this exercise is difficult!!!!!!

Bài 1:Áp dụng C-S dạng engel

\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)

Ta có : \(A=\frac{2019}{x+xy+1}+\frac{2019}{y+yz+1}+\frac{2019}{z+zx+1}=2019\left(\frac{1}{x+xy+1}+\frac{1}{y+yz+1}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+xyz+z}+\frac{xz}{xyz+xyz^2+xz}+\frac{1}{z+zx+1}\right)\)

\(=2019\left(\frac{z}{xz+z+1}+\frac{xz}{1+z+xz}+\frac{1}{z+zx+1}\right)\)(vì xyz = 1)

\(=2019\left(\frac{z+xz+1}{xz+z+1}\right)=2019\)

Vậy A = 2019