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

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

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

\(M=2019\)

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

\(\Rightarrow\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{x+z}+\frac{z\left(x+y+z\right)}{x+y}=x+y+z\)

\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)

\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)

\(\Rightarrow M=2019+0=2019\)

Bài tập đâu rồi?

Vì \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\)

\(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=x+y+z\)

\(\Leftrightarrow\frac{x^2}{y+z}+\frac{xy}{z+x}+\frac{zx}{x+y}+\frac{xy}{y+z}+\frac{y^2}{z+x}+\frac{yz}{x+y}+\frac{zx}{y+z}+\frac{yz}{z+x}+\frac{z^2}{x+y}=x+y+z\)

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

\(\Leftrightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\frac{y\left(z+x\right)}{z+x}+\frac{z\left(x+y\right)}{x+y}+\frac{x\left(y+z\right)}{y+z}=x+y+z\)

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

\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)

\(\Rightarrow M=2019\)

Ta có: \(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=x+y+z\)

TH1: \(x+y+z=0\)

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

\(\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{x+y+z}{x+y-3+y+z+1+z+x+2}\)

                       \(=\frac{x+y+z}{x+y+y+z+z+x}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

\(\Rightarrow x+y=\frac{1}{2}-z\)

      \(y+z=\frac{1}{2}-x\)

      \(z+x=\frac{1}{2}-y\)

Thay \(x+y-3=\frac{1}{2}-z-3\)

\(\Rightarrow\frac{z}{\frac{1}{2}-z+3}=\frac{1}{2}\)

\(\Rightarrow2z=\frac{1}{2}-z-3\)

\(\Rightarrow2z+z=\frac{1}{2}-3\)

\(\Rightarrow3z=-\frac{5}{2}\Rightarrow z=-\frac{5}{6}\)

Thay \(y+z+1=\frac{1}{2}-x+1\)

\(\Rightarrow\frac{x}{\frac{1}{2}-x+1}=\frac{1}{2}\)

\(\Rightarrow2x=\frac{1}{2}-x+1\)

\(\Rightarrow2x+x=\frac{1}{2}+1\)

\(\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

Thay \(z+x+2=\frac{1}{2}-y+2\)

\(\Rightarrow\frac{y}{\frac{1}{2}-y+2}=\frac{1}{2}\)

\(\Rightarrow2y=\frac{1}{2}-y+2\)

\(\Rightarrow2y+y=\frac{1}{2}+2\)

\(\Rightarrow3y=\frac{5}{2}\Rightarrow y=\frac{5}{6}\)

Ta có: \(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)

                \(=\left(\frac{1}{2}+\frac{5}{6}+-\frac{5}{6}-\frac{3}{2}\right)^{2019}\)

                \(=\left[\left(\frac{1}{2}-\frac{3}{2}\right)+\left(-\frac{5}{6}+\frac{5}{6}\right)\right]^{2019}\)

                 \(=\left(-1\right)^{2019}=-1\)

TH2: x + y + z = 0

\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=0\)

\(\Rightarrow x=y=z=0\)

\(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)

    \(=\left(0-\frac{3}{2}\right)^{2019}=\left(-\frac{3}{2}\right)^{2019}\)

Ah! Mk nhầm chút. TH1 là khác 0 nhé!!!!!!

Đặt \(\dfrac{1}{a}=\dfrac{1}{x+y},\dfrac{1}{b}=\dfrac{1}{y+z},\dfrac{1}{c}=\dfrac{1}{z+x}\)

Đề trở thành: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\), tính \(P=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tương đương \(ab+bc=-ac\)

\(P=\dfrac{b^3c^3+a^3c^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(ab+bc\right)\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}=\dfrac{-ac\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}\)

\(=\dfrac{a^2c^2-a^2b^2+ab^2c-b^2c^2}{ab^2c}=\dfrac{ac}{b^2}-\dfrac{a}{c}+1-\dfrac{c}{a}\)\(=ac\left(\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\right)-\dfrac{a}{c}+1-\dfrac{c}{a}\) (do \(\dfrac{1}{b}=-\dfrac{1}{a}-\dfrac{1}{c}\) tương đương \(\dfrac{1}{b^2}=\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\)) 

\(=3\)

Vậy P=3