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

x+y+z hay là xyz hả bạn

x*y*z =2018 nha

câm mồm

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

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

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

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

\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

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

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

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

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

Thay xyz=2013 vào ta có:

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

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

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

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

\(=\frac{xz+1+z}{xz+z+1}=1\) (Đpcm)

Lời giải:

Đặt mẫu số của $B$ là $M$.

Từ \(2018x^3=2019y^3=2020z^3\)

\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)

\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)

\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)

\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)

\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)