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\(\hept{\begin{cases}\frac{x}{y}=\frac{4}{7}\\\frac{y}{z}=\frac{14}{3}\end{cases}}\Rightarrow\frac{x}{z}=\frac{8}{3}\)
\(\Rightarrow\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}=\frac{14}{3}+\frac{8}{3}=\frac{22}{3}\)
\(A=\dfrac{\dfrac{1}{9}:\dfrac{7}{5}:\dfrac{4}{3}}{\dfrac{1}{81}:\dfrac{49}{25}:\dfrac{16}{9}}=\dfrac{5}{84}:\dfrac{25}{7056}=\dfrac{84}{5}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2013}=\frac{1}{x+y+z}\Rightarrow\frac{yz+xz+xy}{xyz}=\frac{1}{x+y+z}\Rightarrow\left(yz+xz+xy\right)\left(x+y+z\right)=xyz\)
\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz+xyz=xyz\)
\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz=0\)
\(\Rightarrow\left(x^2y+x^2z+xy^2+xyz\right)+\left(y^2z+xz^2+y^2z+xyz\right)=0\)
\(\Rightarrow x\left(xy+xz+y^2+yz\right)+z\left(yz+xz+y^2+xy\right)=0\)
\(\Rightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+y=0\Rightarrow x^3+y^3=0\\y+z=0\Rightarrow y^5+z^5=0\\x+z=0\Rightarrow z^7+x^7=0\end{cases}}\)
\(\Rightarrow A=\left(x^3+y^3\right)\left(y^5+z^5\right)\left(z^7+x^7\right)=0\)
Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)
\(\orbr{\begin{cases}y=\frac{3}{x}\\z=\frac{4}{x}\end{cases}\Rightarrow\frac{12}{x^2}=6\Rightarrow x^2=2}\)
\(\orbr{\begin{cases}x=\frac{3}{y}\\z=\frac{6}{y}\end{cases}\Rightarrow\frac{18}{y^2}=4\Rightarrow y^2=\frac{9}{2}}\)
\(\orbr{\begin{cases}x=\frac{4}{z}\\y=\frac{6}{z}\end{cases}\Rightarrow\frac{24}{z^2}=3\Rightarrow z^2=8}\)
\(A=\frac{1}{2}\left(2+\frac{9}{2}+8\right)=\frac{4+9+16}{4}=\frac{29}{4}\)
Ta có:
\(\dfrac{x}{y}=\dfrac{4}{7}\Rightarrow x=\dfrac{4}{7}y\)
Mà \(\dfrac{y}{z}=\dfrac{14}{3}\Rightarrow y=\dfrac{14}{3}z\)
Nên \(x=\dfrac{4}{7}y=\dfrac{4}{7}.\dfrac{14}{3}z=\dfrac{8}{3}z\)
Ta có:
\(\dfrac{x+y}{z}=\dfrac{\dfrac{14}{3}z+\dfrac{8}{3}z}{z}=\dfrac{\dfrac{22}{3}z}{z}=\dfrac{22}{3}\)