Từ \(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
Khi đó \(P=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5\cdot\left(3k\right)^2+3\cdot\left(5k\right)^2}{10\cdot\left(3k\right)^2-3\cdot\left(5k\right)^2}\)
\(=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=\dfrac{45k^2+75k^2}{90k^2-75k^2}\)
\(=\dfrac{120k^2}{15k^2}=\dfrac{120}{15}=8\)
Ta có:
x/3=y/5
=> x=3/5y
Thay x vào P ta được P
\(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
Ta có:
\(P=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)
\(=\dfrac{45k^2+75k^2}{90k^2-75k^2}=\dfrac{15k^2\left(3+5\right)}{15k^2\left(6-5\right)}=\dfrac{3+5}{6-5}=8\)
Vậy \(P=8\)
Ta có:
\(\dfrac{x}{y}=\dfrac{3}{5}\)=> \(\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}\)\(=k\)
=> \(x=3k\)
\(y=5k\)
Khi đó: \(P=\dfrac{5.x^2+3.y^2}{10.x^2-3.y^2}\)=\(\dfrac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)=\(\dfrac{45.k^2+75.k^2}{90.k^2-75.k^2}\)=\(\dfrac{120.k^2}{15.k^2}\)=\(\dfrac{120}{15}\)=\(8\)
Vậy: Gía trị của biểu thức \(P=\dfrac{5.x^2+3.y^2}{10.x^2-3.y^2}\) tại \(\dfrac{x}{y}=\dfrac{3}{5}\) là \(8\)
