\(\frac{x}{3}=\frac{y}{5}\)\(\Rightarrow x=\frac{3y}{5}\)
Thay vào biểu thức A ta được:
\(A=\frac{5.\left(\frac{3y}{5}\right)^2+3y^2}{10.\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\frac{9y^2+15y^2}{5}}{\frac{18y^2-15y^2}{5}}=\frac{24y^2}{3y^2}=8\)
Đặt \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow x=3k,y=5k\)
Ta có: \(A=\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5.\left(3k\right)^2+3.\left(5k\right)^2}{10.\left(3k\right)^2-3.\left(5k\right)^2}=\frac{45k^2+75k^2}{90k^2-75k^2}=\frac{k^2\left(45+75\right)}{k^2\left(90-75\right)}=\frac{120k^2}{15k^2}=8\)
x/3 = y/5 => y = 5/3.x => y^2 = 25/9.x^2 => 3y^2 = 25/3.y^2
=> A =(5x^2+25/3.x^2)/(10x^2-25/3.x^2) = 40/3.x^2 / 5/3.x^2 = 8
Vậy A = 8
Tk mk nha
A=5x^2+3y^2/10x^2-3y^2 (1)
Đặt x/3=y/5=k
=>x=3.k
=>y=5.k
Thay vào biểu thức (1):
A=5.(3k)^2+3.(5k)^2/10.(3k)^2-3.(5k)^2
=45k^2+75k^2/90k^2-75k^2
=120k^2/15k^2
=8
