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Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\) (*)
Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)
\(=\frac{x}{z}+\frac{y}{z}+\frac{x}{y}+\frac{x}{y}+\frac{y}{x}+\frac{z}{x}\)
\(=\left(\frac{x}{z}+\frac{x}{y}\right)+\left(\frac{y}{x}+\frac{y}{z}\right)+\left(\frac{z}{x}+\frac{z}{y}\right)\)
\(=x\left(\frac{1}{z}+\frac{1}{y}\right)+y\left(\frac{1}{x}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)
Thay (*) vào,ta có : \(A=x.\left(\frac{-1}{x}\right)+y.\left(-\frac{1}{y}\right)+z.\left(-\frac{1}{z}\right)=\left(-1\right)+\left(-1\right)+\left(-1\right)=-3\)
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\Rightarrow\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{x+z}+\frac{z\left(x+y+z\right)}{x+y}=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
\(\Rightarrow M=2019+0=2019\)
...
=>\(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=1\)
=>\(\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{xy}{z+x}+\frac{y^2}{z+x}+\frac{yz}{z+x}+\frac{xz}{x+y}+\frac{yz}{x+y}+\frac{z^2}{x+y}=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{xy}{z+x}+\frac{yz}{z+x}+\frac{xz}{x+y}+\frac{yz}{x+y}\right)=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy+xz}{y+z}+\frac{xy+yz}{z+x}+\frac{xz+yz}{x+y}\right)=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(x+y+z\right)=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+1=1\)
=>\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
Từ \(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\Rightarrow\frac{x}{y-z}=-\frac{y}{z-x}-\frac{z}{x-y}\)
\(\Rightarrow\frac{x}{y-z}=\frac{y}{x-z}+\frac{z}{y-x}\)
\(\Leftrightarrow\frac{x}{y-z}=\frac{y\left(y-x\right)+z\left(x-z\right)}{\left(x-z\right)\left(y-x\right)}\)
\(\Leftrightarrow\frac{x}{y-z}=\frac{y^2-xy+zx-z^2}{\left(x-z\right)\left(y-x\right)}\)
\(\Leftrightarrow\frac{x}{\left(y-z\right)^2}=\frac{y^2-xy+zx-z^2}{\left(x-z\right)\left(y-x\right)\left(y-z\right)}\)
C/m tương tự đc \(\frac{y}{\left(z-x\right)^2}=\frac{z^2-yz+xy-x^2}{\left(x-z\right)\left(y-z\right)\left(y-z\right)}\)
\(\frac{z}{\left(x-y\right)^2}=\frac{x^2-xz+zy-y^2}{\left(x-z\right)\left(y-x\right)\left(y-z\right)}\)
Khi đó \(Q=\frac{y^2-xy+xz-z^2+z^2-yz+xy-x^2+x^2-xz+yz-y^2}{\left(x-z\right)\left(y-x\right)\left(y-z\right)}=0\)
Vậy Q=0
\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{x+y+z}{z}-1+\frac{x+y+z}{y}-1+\frac{x+y+z}{x}-1\)
\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3=0-3=-3\)
Vì \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{xy}{z+x}+\frac{zx}{x+y}+\frac{xy}{y+z}+\frac{y^2}{z+x}+\frac{yz}{x+y}+\frac{zx}{y+z}+\frac{yz}{z+x}+\frac{z^2}{x+y}=x+y+z\)
\(\Leftrightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy+yz}{z+x}\right)+\left(\frac{yz+zx}{x+y}\right)+\left(\frac{zx+xy}{y+z}\right)=x+y+z\)
\(\Leftrightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\frac{y\left(z+x\right)}{z+x}+\frac{z\left(x+y\right)}{x+y}+\frac{x\left(y+z\right)}{y+z}=x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+x+y+z=x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
\(\Rightarrow M=2019\)
1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)
\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm
2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)
tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1
3) kiểm tra lại xem đề đã chuẩn chưa