Ta có: \(tan\alpha=-2\)
=>\(\dfrac{sin\alpha}{cos\alpha}=-2\)
=>\(sin\alpha=-2\cdot cos\alpha\)
\(P=\dfrac{3\cdot cos\alpha+4\cdot sin\alpha}{cos\alpha+sin\alpha}=\dfrac{3cos\alpha+4\cdot\left(-2cos\alpha\right)}{cos\alpha-2\cdot cos\alpha}\)
\(=\dfrac{3+4\cdot\left(-2\right)}{1-2}=\dfrac{3-8}{-1}=5\)