Question

Cho 4x = 3y và 5y = 6z

Tính: \(A=\frac{2x^2-3y^2-4z^2}{4xy-3yz+2xz}\)

Answer 1

Giải:

Ta có:

\(4x=3y\Rightarrow\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\)

\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)

Đặt \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=k\)

\(\Rightarrow x=9k,y=12k,z=10k\)

Ta có:
\(A=\frac{2x^2-3y^2-4z^2}{4xy-3yz+2xz}\)

\(\Rightarrow A=\frac{2\left(9k\right)^2-3\left(12k\right)^2-4\left(10k\right)^2}{4.9.k.12.k-3.12.k.10.k+2.9.k.10k}\)

\(\Rightarrow A=\frac{2.9.k^2-3.12.k^2-4.10.k^2}{432.k^2-360.k^2+180.k^2}\)

\(\Rightarrow A=\frac{18.k^2-36.k^2-40.k^2}{k^2.\left(432-360+180\right)}\)

\(\Rightarrow A=\frac{k^2.\left(18-36-40\right)}{k^2.252}\)

\(\Rightarrow A=\frac{-58}{252}\)

\(\Rightarrow A=\frac{-1}{3}\)

Vậy \(A=\frac{-1}{3}\)

 

Answer 2

Ta có:

\(\begin{cases}4x=3y\\5y=6z\end{cases}\) => \(\begin{cases}\frac{x}{3}=\frac{y}{4}\\\frac{y}{6}=\frac{z}{5}\end{cases}\)=> \(\begin{cases}\frac{x}{9}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{10}\end{cases}\) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)

Đặt \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=k\)

=> \(\begin{cases}x=9k\\y=12k\\z=10k\end{cases}\)

Ta có:

\(A=\frac{2.\left(9k\right)^2-3.\left(12k\right)^2-4.\left(10k\right)^2}{4.9k.12k-3.12k.10k+2.9k.10k}\)

\(A=\frac{2.81.k^2-3.144.k^2-4.100.k^2}{432k^2-360k^2+180k^2}\)

\(A=\frac{162k^2-432k^2-400k^2}{252k^2}\)

\(A=\frac{-670k^2}{252k^2}=\frac{-335}{126}\)