Cho \(tan\alpha =\frac{1}{2}(\beta \in (0;\frac{\pi}{2}))\). Tính giá trị biểu thức: \(P=\frac{2sin\frac{\pi}{2}+3cos\frac{\alpha }{2}}{sin\frac{\alpha }{2}+2cos\frac{\alpha }{2}}+\frac{1}{\sqrt{5}}\)
Vì \(tan\alpha =\frac{1}{2}(\alpha \in (0;\frac{\pi}{2}))\) nên \(\frac{2tan\frac{\alpha }{2}}{1-tan^2\frac{\alpha }{2}}=\frac{1}{2}\Leftrightarrow tan^2\frac{\alpha }{2}+4tan\frac{\alpha }{2}-1=0\) Suy ra \(tan\frac{\alpha }{2}=-2+\sqrt{5}\) hoặc \(tan\frac{\alpha }{2}=-2-\sqrt{5} (l)\). Do \(tan\frac{\alpha }{2}>0\) Thay vào ta có \(P=\frac{2tan\frac{\alpha }{2}+3}{tan\frac{\alpha }{2}+2}+\frac{1}{\sqrt{5}}=\frac{2\sqrt{5}-1}{\sqrt{5}}+\frac{1}{\sqrt{5}}=2\)