Question

Cho \(\frac{a}{b}=\frac{3}{4}\).Tính \(A=\frac{a^2+b^2}{a^2-b^2}\)

Answer 1

tìm a, b

Answer 2

 \(\frac{a}{b}=\frac{3}{4}\) \(< =>\frac{a}{3}=\frac{b}{4}\)

Đặt \(\frac{a}{3}=\frac{b}{4}=k=>a=3k;b=4k\)

Ta có: \(A=\frac{a^2+b^2}{a^2-b^2}=\frac{\left(3k\right)^2+\left(4k\right)^2}{\left(3k\right)^2-\left(4k\right)^2}=\frac{9k^2+16k^2}{9k^2-16k^2}=\frac{\left(9+16\right).k^2}{\left(9-16\right).k^2}=\frac{28k^2}{-7k^2}=\frac{28}{-7}=-4\)

Vậy A=-4
 

Answer 3

cho sửa lại nhé!

kết quả bằng \(\frac{25}{-7}\)chứ ko phải -4