Question

cho\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\). tính:

A=\(\dfrac{5x^2+3y^2}{10x^2-3y^2}\)

Answer 1

Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\) (k \(\ne\) 0)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)

Mà A = \(\dfrac{5x^2+3y^2}{10x^2-3y^2}\) (bài cho)

\(\Rightarrow\) A = \(\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\)

\(\Leftrightarrow\) A = \(\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}\)

\(\Leftrightarrow\) A = \(\dfrac{45k^2+75k^2}{90k^2-75k^2}\)

\(\Leftrightarrow\) A = \(\dfrac{120k^2}{15k^2}\)

\(\Leftrightarrow\) A = \(\dfrac{120}{15}\)

\(\Leftrightarrow\) A = 8

Vậy A = 8