+) ta có : \(A=\dfrac{sinx-cosx}{2sinx+cosx}=\dfrac{\dfrac{sinx}{sinx}-\dfrac{cosx}{sinx}}{\dfrac{2sinx}{sinx}+\dfrac{cosx}{sinx}}\) \(=\dfrac{1-cotx}{2+cotx}\)
\(=\dfrac{1-3\sqrt{8}}{2+3\sqrt{8}}=\dfrac{-37+9\sqrt{2}}{34}\)
+) ta có : \(A=\dfrac{1+sin^2x}{2+sinx.cosx}=\dfrac{sin^2x+cos^2x+sin^2x}{2sin^2x+2cos^2x+sinx.cosx}\) \(=\dfrac{2sin^2x+cos^2x}{2sin^2x+2cos^2x+sinx.cosx}=\dfrac{\dfrac{2sin^2x}{sin^2x}+\dfrac{cos^2x}{sin^2x}}{\dfrac{2sin^2x}{sin^2x}+\dfrac{2cos^2x}{sin^2x}+\dfrac{sinx.cosx}{sin^2x}}\) \(=\dfrac{2+cot^2x}{2+2cot^2x+cotx}=\dfrac{2+\left(3\sqrt{8}\right)^2}{2+2\left(3\sqrt{8}\right)^2+3\sqrt{8}}=\dfrac{74}{146+3\sqrt{8}}\)
