Question

Cho \(\frac{x}{3}=\frac{y}{5}\)Tính \(C=\frac{5x^2+3y^2}{10x^2-3y^2}\)

Answer 1

Đặt \(\frac{x}{3}=\frac{y}{5}=k\)=> \(x=3k\) ; \(y=5k\)

Khi đó, ta có: C = \(\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}\)

                          = \(\frac{5.3^2.k^2+3.5^2.k^2}{10.3^2.k^2-3.5^2.k^2}\)

                          = \(\frac{k^2.\left(5.9+3.25\right)}{k^2.\left(10.9-3.25\right)}\)

                          =  8

Answer 2

\(\frac{x}{3}=\frac{y}{5}\Rightarrow5x=3y\)

\(C=\frac{3xy+5xy}{6xy-5xy}=\frac{8xy}{1xy}=8\)

cách này nhanh hơn không :v