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\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
\(A=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{y+z}{z}\)
Do \(x-y-z=0\)
\(\Rightarrow x-z=y;y-x=-z;y+z=x\)
Khi đó \(A=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
Vậy A=-1
\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)
\(=\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{1+yz+y}\)
\(=\frac{1}{xy+x+1}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz}{xy\cdot yz+xyz+yz}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz}{yz+y+1}+\frac{y+1}{yz+y+1}\)
\(=\frac{yz+y+1}{yz+y+1}\)
\(=1\)
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Ta có: \(\frac{a}{x}+\frac{y}{b}=1\)
\(\rightarrow\frac{a}{x}\cdot\frac{b}{y}+\frac{y}{b}\cdot\frac{b}{y}=1\cdot\frac{b}{y}\)
\(\rightarrow\frac{ab}{xy}+1=\frac{b}{y}\left(1\right)\)
Ta có: \(\frac{b}{y}+\frac{z}{c}=1\)
\(\rightarrow\frac{b}{y}=1-\frac{z}{c}\left(2\right)\)
Từ (1) và (2) \(\rightarrow\frac{ab}{xy}+1=1-\frac{z}{c}\)
\(\rightarrow\frac{ab}{xy}=\frac{-z}{c}\) \(\rightarrow abc=-xyz\)
\(\rightarrow abc+xyz=0\)
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Ta có : x/a+y/b = xy/ay+ ba/ay=xy+ab/ay=1
=>xy+ab=ay
=>xy=a(y-b)
=>y-b=xy/a (1)
Ta lại có :y/b+c/z=yz/bz +bc/bz=yz+bc/bz=1
=>yz+bc=bz
=>yz-bz=-bc
=>z(y-b)=-bc
=>y-b=-bc/z (2)
Từ (1),(2)=> xy/a=-bc/z( =y-b) => xyz= -abc
=>xyz+abc=0
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a)Ta có : 2x+2y-z-7=0 => 2x+2y-z=7
Ta có : \(x=\frac{y}{2}=>\frac{x}{2}=\frac{y}{4}\)
Mà \(\frac{y}{4}=\frac{z}{5}\)nên \(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}=\frac{2x+2y-z}{4+8-5}=\frac{7}{7}=1\)
Từ \(\frac{x}{2}=1=>x=2\)
Từ\(\frac{y}{4}=1=>y=4\)
Từ \(\frac{z}{5}=1=>z=5\)
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)
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