Ta có:

\(\frac{x}{x+1}=1-\frac{1}{x+1}\)

\(\frac{y}{y+1}=1-\frac{y}{y+1}\)

\(\frac{z}{z+4}=1-\frac{4}{z+4}\)

\(\Rightarrow\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\)

\(\le\left[3-\left(\frac{4}{x+y+2}+\frac{4}{z+4}\right)\right]\le\left(3-\frac{16}{x+y+z+6}\right)=3-\frac{16}{6}=\frac{1}{3}\)

Ta có: \(x+y+z=0\)

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

Ta có: \(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

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

\(=\dfrac{-z}{y}\cdot\dfrac{-x}{z}\cdot\dfrac{-y}{x}\)

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

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Đề sai rồi phãi là: \(A= ( 1-\frac{z}{x})(1-\frac{x}{y})(1-\frac{y}{z}) \)

\(A=\left(\frac{x-z}{x}\right)\left(\frac{y-x}{y}\right)\left(\frac{z-y}{z}\right)\)

Từ x-y-z = 0 \(\Rightarrow\) x-z = y

y-x = z

z-y = x

Thay vào A, ta có: \(\left(\frac{y}{x}\right)\left(\frac{z}{y}\right)\left(\frac{x}{z}\right)\)

\(\Rightarrow A = 1 \)

Bạn Lê Anh Tú xem lại đi \(z-y\ne x\)

\(B=\left(\frac{x-z}{x}\right).\left(\frac{y-x}{y}\right).\left(\frac{y+z}{z}\right)\)

Từ x-y-z=0 \(\Rightarrow x-z=y\)

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

\(\Rightarrow y+z=x\)

Thay vào B ta được

\(B=\left(\frac{y}{x}\right).\left(\frac{-z}{y}\right).\left(\frac{x}{z}\right)=-1\)