Question

[Lê Ngọc Linh]

Tính giá trị biểu thức A=\(\frac{5x^2+3y^2}{10x^2-3y^2}\)với \(\frac{x}{3}=\frac{y}{5}\)

Answer 1

[nhung nguyễn]

Đặt \(\frac{x}{3}=\frac{y}{5}=n\Rightarrow x=3n;y=5n\)

\(\Rightarrow A=\frac{5.3^2n^2+3.5^2n^2}{10.3^2n^2-3.5^2n^2}=\frac{n^2\left(45+75\right)}{n^2\left(90-75\right)}=\frac{n^2.120}{n^2.25}=\frac{24}{5}\)

Answer 2

[Lê Minh Anh]

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

Thay 3y = 5x ; ta được: 

\(A=\frac{5x^2+5x^2}{10x^2-5x^2}=\frac{2\times5x^2}{2\times5x^2-5x^2}=\frac{2\times5x^2}{5x^2\times\left(2-1\right)}=\frac{2\times5x^2}{5x^2\times1}=2\)