Question

cho các số nguyên dương a , b thỏa mãn 2a ^2- b^2 / a^2+b^2=-1/13. Tìm dạng tối giản của a/b

Answer 1

\(\dfrac{2a^2-b^2}{a^2+b^2}=-\dfrac{1}{13}\)

\(\Leftrightarrow\dfrac{\left(2a^2+2b^2\right)-3b^2}{a^2+b^2}=-\dfrac{1}{13}\)

\(\Leftrightarrow2-\dfrac{3b^2}{a^2+b^2}=-\dfrac{1}{13}\)

\(\Leftrightarrow\dfrac{b^2}{a^2+b^2}=\dfrac{9}{13}\)

\(\Rightarrow1-\dfrac{b^2}{a^2+b^2}=1-\dfrac{9}{13}=\dfrac{4}{13}\)

\(\Leftrightarrow\dfrac{a^2}{a^2+b^2}=\dfrac{4}{13}\)

 

\(\dfrac{a^2}{b^2}=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{2}{3}\\\dfrac{a}{b}=-\dfrac{2}{3}\end{matrix}\right.\)