trình bày chi tiết hai bài này giúp em vlwis sj em cảm ơn

Câu 17:

Ta có: $\cot \alpha = \dfrac{\cos \alpha}{\sin \alpha} = -3 \Rightarrow \cos \alpha = -3\sin \alpha$

$P = \left| \dfrac{\sin^3 \alpha + \cos^3 \alpha}{\sin \alpha - \cos \alpha} \right| = \left| \dfrac{\sin^3 \alpha + (-3\sin \alpha)^3}{\sin \alpha - (-3\sin \alpha)} \right| = \left| \dfrac{-26\sin^3 \alpha}{4\sin \alpha} \right| = \dfrac{13}{2}\sin^2 \alpha$

Lại có: $\sin^2 \alpha = \dfrac{1}{1 + \cot^2 \alpha} = \dfrac{1}{1 + (-3)^2} = \dfrac{1}{10} = 0{,}1$

$\Rightarrow P = \dfrac{13}{2} \cdot 0{,}1 = 0{,}65$

Câu 18:

Ta có: $\text{BC}^2 = \text{AB}^2 + \text{AC}^2 - 2\text{AB} \cdot \text{AC} \cdot \cos\widehat{\text{A}}$

$\Rightarrow \text{BC}^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^\circ = 9 + 25 - 30 \cdot \left(-\dfrac{1}{2}\right) = 49$

$\Rightarrow \text{BC} = 7$

$\cos\widehat{\text{ABC}} = \dfrac{\text{AB}^2 + \text{BC}^2 - \text{AC}^2}{2\text{AB} \cdot \text{BC}} = \dfrac{3^2 + 7^2 - 5^2}{2 \cdot 3 \cdot 7} = \dfrac{11}{14}$

Mà $\widehat{\text{ABD}} = 180^\circ - \widehat{\text{ABC}}$

$\Rightarrow \cos\widehat{\text{ABD}} = -\cos\widehat{\text{ABC}} = -\dfrac{11}{14}$

Lại có: $\text{AD}^2 = \text{AB}^2 + \text{BD}^2 - 2\text{AB} \cdot \text{BD} \cdot \cos\widehat{\text{ABD}}$

$\Rightarrow \text{AD}^2 = 3^2 + 7^2 - 2 \cdot 3 \cdot 7 \cdot \left(-\dfrac{11}{14}\right) = 9 + 49 + 33 = 91$

$\Rightarrow \text{AD} = \sqrt{91} \approx 9{,}5$