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= \(\dfrac{\sqrt{xy}-1+\sqrt{yz}-3+\sqrt{zx}-5}{3+9+6}\) = \(\dfrac{11-\left(1+3+5\right)}{18}\)=\(\dfrac{1}{9}\) 
\(\dfrac{1}{xy+x+1}+\dfrac{y}{yz+y+1}+\dfrac{1}{xyz+yz+y}\)
\(=\dfrac{xyz}{xy+x+xyz}+\dfrac{y}{yz+y+1}+\dfrac{1}{yz+y+1}\)
\(=\dfrac{xyz}{x\left(y+1+yz\right)}+\dfrac{y}{yz+y+1}+\dfrac{1}{yz+y+1}\)
\(=\dfrac{yz}{yz+y+1}+\dfrac{y}{yz+y+1}+\dfrac{1}{yz+y+1}\)
\(=\dfrac{yz+y+1}{yz+y+1}=1\left(đpcm\right)\)
Vậy...
êu , sao \(\dfrac{1}{xy+x+1}\)+... lại bằng \(\dfrac{xyz}{xy+z+zxy}\)+... vậy ?
\(A=\dfrac{x}{xy+x+1}+\dfrac{y}{y+1+yz}+\dfrac{z}{1+z+xz}\)
\(=\dfrac{x}{xy+x+xyz}+\dfrac{y}{y+1+yz}+\dfrac{yz}{1+yz+z}\)
\(=\dfrac{x}{x\left(y+1+yz\right)}+\dfrac{y}{y+1+yz}+\dfrac{yz}{1+yz+y}\)
\(=\dfrac{1}{y+1+yz}+\dfrac{y}{y+1+yz}+\dfrac{yz}{1+yz+y}\)
\(=\dfrac{1+y+yz}{y+1+yz}=1.\)
\(\dfrac{xy}{x+y}=\dfrac{yz}{y+z}=\dfrac{zx}{z+x}\\ \Rightarrow\dfrac{x+y}{xy}=\dfrac{y+z}{yz}=\dfrac{z+x}{zx}\\ \Rightarrow\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{1}{x}+\dfrac{1}{z}\\ \Rightarrow\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}\\ \Rightarrow x=y=z\)
\(\Rightarrow P=\dfrac{xy+yz+zx}{x^2+y^2+z^2}=\dfrac{x^2+x^2+x^2}{x^2+x^2+x^2}=1\)
a) Ta có:
\(x+y+z=49\Rightarrow12x+12y+12z=588\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{588}{49}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.3:2\\y=12.4:3\\z=12.5:4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{yz+xz+xy}{xyz}=0\)
\(\Leftrightarrow xy+xz+yz=0\)
~ ~ ~
\(x^2+2yz\)
\(=x^2+yz-xy-xz\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự, ta có: \(y^2+2xz=\left(y-x\right)\left(y-z\right)\) và \(z^2+2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{\left(x-z\right)\left(x-y\right)}+\dfrac{xz}{\left(y-x\right)\left(y-z\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(A=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)
\(=\dfrac{\left(x-z\right)\left(x-y\right)\left(y-z\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)
= 1
\(A=\dfrac{5}{x+xy+xyz}+\dfrac{5x}{xy+xyz+x}+\dfrac{5xy}{xyz+x.xyz+xy}\)
Vì \(xyz=1\)
\(\Rightarrow A=\dfrac{5}{x+xy+xyz}+\dfrac{5x}{xy+xyz+x}+\dfrac{5xy}{xyz+x+xy}\)
\(\Rightarrow A=5.\dfrac{x+xy+xyz}{x+xy+xyz}=5\)