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Ta có:
\(x:y:z=5:4:3\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=3k\end{matrix}\right.\)
\(\Rightarrow\frac{2x+3y-5z}{2x-3y+5z}=\frac{2.5k+3.4k-5.3k}{2.5k-3.4k+5.3k}=\frac{10k+12k-15k}{10k-12k+15k}=\frac{7k}{13k}=\frac{7}{13}\)
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{-4k-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-1k}=-16\)
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow x=-4k;y=-7k;z=3k\) (1)
Thay (1) vào A , ta được
\(A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{2\left(-4k\right)-3\left(-7k\right)-6.3k}\)
\(\Rightarrow A=\dfrac{8k+\left(-7k\right)+15k}{-8k+21k+\left(-18k\right)}\)
\(\Rightarrow A=\dfrac{k[8+\left(-7\right)+15]}{k[-8+21+\left(-18\right)]}\)
\(\Rightarrow A=\dfrac{16k}{-5k}\)
\(\Rightarrow A=\dfrac{16}{5}\)
Vậy \(A=\dfrac{16}{5}\)
Ta có : \(\dfrac{x}{y}=\dfrac{3}{4};\dfrac{y}{z}=\dfrac{5}{6}\)
\(\Rightarrow4x=3y;6y=5z\)
\(\Leftrightarrow8x=6y=5z.\)
\(\Rightarrow y=\dfrac{8x}{6}=\dfrac{4}{3}x\)
Thay vào A ta có :
\(A=\dfrac{2x+3y+5z}{y+5z}=\dfrac{2x+4x+8x}{\dfrac{4}{3}x+8x}=\dfrac{3}{2}.\)
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2\left(-4k\right)-7k+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-5k}=-\dfrac{16}{5}\)
\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{2x+3y-5z}{10+12-15}=\dfrac{2x-3y+5z}{10-12+15}\\ \Rightarrow A=\dfrac{10+12-15}{10-12+15}=\dfrac{7}{13}\)
A=7/13