Đặt $\frac{a-b}{a+b}$ = A
=> A² = $\frac{a-b}{a+b}$ = $\frac{(a-b)^{2}}{(a+b)^2}$
= $\frac{a² - 2ab + b²}{a^2 + 2ab + b^2}$ = $\frac{3a² - 6ab + 3b²}{3a^2 + 6ab + 3b^2}$
= $\frac{10ab - 6ab}{ 10ab + 6ab}$ = $\frac{4ab}{16ab}$ = $\frac{1}{4}$
=> A = 1/2