Question

cho 12x²+13y²=25xy . Tính      \(A=\frac{x+y}{x-y}\)

Answer 1

\(12x^2+13y^2=25xy\)

\(\Leftrightarrow12x^2-25xy+13y^2=0\)

\(\Leftrightarrow12x^2-12xy-13xy+13y^2=0\)

\(\Leftrightarrow12x\left(x-y\right)-13y\left(x-y\right)=0\)

\(\Leftrightarrow\left(12x-13y\right)\left(x-y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}12x-13y=0\\x-y=0\end{cases}}\)

Mà để A xác định \(\Leftrightarrow x-y\ne0\) Do đó \(12x-13y=0\Leftrightarrow12x=13y\Rightarrow x=\frac{13}{12}y\)

\(\Rightarrow A=\frac{\frac{13}{12}y+y}{\frac{13}{12}y-y}=\frac{y\left(\frac{13}{12}+1\right)}{y\left(\frac{13}{12}-1\right)}=\left(\frac{13}{12}+1\right):\left(\frac{13}{12}-1\right)=\frac{25}{12}:\frac{1}{12}=25\)