TH1: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(y+z=-x\)

\(x+z=-y\)

\(\Rightarrow M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\dfrac{-xyz}{8xyz}=\dfrac{-1}{8}\)

TH2: \(x+y+z\ne0\)

\(\Rightarrow2x+2y-z=3\)

\(\Rightarrow2x+2y=4z\)

\(\Rightarrow x+y=2z\)

\(x+z=2y\)

\(y+z=2x\)

\(\Rightarrow M=\dfrac{2z.2y.2x}{8xyz}=1\)

Vậy: \(M=\dfrac{-1}{8}\) hoặc \(1\)

Ta có \(\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\Rightarrow\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x+2y-z}{z}=3\\\dfrac{2x+2z-y}{y}=3\\\dfrac{2y+2z-x}{x}=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2y-z=3z\\2x+2z-y=3y\\2y+2z-x=3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2y=4z\\2x+2z=4y\\2y+2z=4x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=2z\\x+z=2y\\y+z=2x\end{matrix}\right.\)

Ta có \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}\)

\(\Rightarrow M=\dfrac{2x.2y.2z}{8xyz}=\dfrac{8xyz}{8xyz}=1\)

Vậy \(M=1\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x+2y-z}{z}=\dfrac{2x-y+2z}{y}=\dfrac{-x+2y+2z}{x}=\dfrac{2x+2y-z+2x-y+2z-x+2y+2x}{x+y+z}=\dfrac{3x+3y+3z}{x+y+z}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)

\(\Rightarrow\)\(\dfrac{2x+2y-z}{z}=3\Leftrightarrow2x+2y-z=3z\Leftrightarrow2\left(x+y\right)=4z\Leftrightarrow x+y=2z\Leftrightarrow z=\dfrac{x+y}{2}\)

Tương tự: \(x=\dfrac{y+z}{2}\)

\(y=\dfrac{x+z}{2}\)

Thay vào M, ta được:

\(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\left(\dfrac{y+z}{2}.\dfrac{x+z}{2}.\dfrac{x+y}{2}\right).8}\)

\(=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8}.8}=1\)

TH1 : \(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)

\(\Leftrightarrow M=\dfrac{\left(-z\right)\left(-x\right)\left(-y\right)}{8xyz}=\dfrac{-\left(xyz\right)}{8xyz}=\dfrac{-1}{8}\)

Th2 : \(x+y+z\ne0\)

\(\dfrac{2x+2y-z}{z}=\dfrac{2x-2z+y}{y}=\dfrac{2y+2z-x}{x}\)

\(\Leftrightarrow\left(\dfrac{2x+2y-z}{z}+3\right)=\left(\dfrac{2x-2z+y}{y}+3\right)=\left(\dfrac{2y+2z-x}{x}+3\right)\)

\(\Leftrightarrow\dfrac{2x+2y+2z}{z}=\dfrac{2x+2y+2z}{y}=\dfrac{2x+2y+2z}{x}\)

\(\Leftrightarrow x=y=z\)

\(\Leftrightarrow M=\dfrac{2x.2y.2z}{8xyz}=1\)

Vậy \(\left[{}\begin{matrix}M=\dfrac{-1}{8}\Leftrightarrow x+y+z=0\\M=1\Leftrightarrow x+y+z\ne0\end{matrix}\right.\)

Biểu thức này chỉ có min, không có max

\(x^2+y^2+z^2=xy+xz+10yz\)

\(\Leftrightarrow\dfrac{3x^2}{4}+\left(\dfrac{x}{2}-y-z\right)^2=12yz\)

\(\Rightarrow12yz\ge\dfrac{3}{4}x^2\Rightarrow yz\ge\dfrac{x^2}{16}\)

\(\Rightarrow P\ge\dfrac{x^3}{2}-\dfrac{3x^3}{2yz}\ge\dfrac{x^3}{2}-\dfrac{3x^3}{\dfrac{x^2}{8}}=\dfrac{x^3}{2}-24x\)

Xét hàm \(f\left(x\right)=\dfrac{x^3}{2}-24x\) với \(x>0\Rightarrow f'\left(x\right)=\dfrac{3}{2}x^2-24=0\Rightarrow x=4\)

Từ BBT ta thấy \(\min\limits_{x>0}f\left(x\right)=f\left(4\right)=-64\)

\(\Rightarrow P_{min}=-64\) khi \(\left(x;y;z\right)=\left(4;1;1\right)\)

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

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

\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)

a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

=0

c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{1}{xyz}\)