Question

cho \(\dfrac{3a^2-b^2}{a^2+b^2}\) = \(\dfrac{3}{4}\). Tính \(\dfrac{a}{b}\)

Answer 1

\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)

\(\Leftrightarrow9a^2=7b^2\)

\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)

hay \(\dfrac{a}{b}\in\left\{\dfrac{\sqrt{7}}{3};-\dfrac{\sqrt{7}}{3}\right\}\)

Answer 2

\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)

\(\Leftrightarrow4.\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)

\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)

\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)

\(\Leftrightarrow9a^2=7b^2\)

\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)

\(\text{hoặc }\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)