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\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)
<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)
<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)
đặt x+y+z = t
=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)
=> x=y=z
ta lại có
\(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)
vì x=y=z => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
\(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)
\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)
\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)
\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)
\(P=\left(\dfrac{x+2y}{y}\right)\left(\dfrac{y+2z}{z}\right)\left(\dfrac{z+2x}{x}\right)\)
Ta có
\(\dfrac{x+2y-z}{z}=\dfrac{y+2z-x}{x}=\dfrac{z+2x-y}{y}=\)
\(=\dfrac{x+2y-z+y+2z-x+z+2x-y}{x+y+z}=\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\dfrac{x+2y}{z}-1=\dfrac{y+2x}{x}-1=\dfrac{z+2x}{y}-1=2\)
\(\Rightarrow\dfrac{x+2y}{z}=\dfrac{y+2x}{x}=\dfrac{z+2x}{y}=3\)
\(\Rightarrow P=3.3.3=27\)
Ta có: \(x-y-z=0\)
\(\Rightarrow x-y=z\)
\(x-z=y\)
\(y+z=x\)
\(\Rightarrow B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{x-z}{x}.\dfrac{-\left(y-x\right)}{y}.\dfrac{z+y}{z}\)
\(=\dfrac{y}{x}.-\dfrac{z}{y}.\dfrac{z}{x}=-1\)
\(\Rightarrow B=-1\)