Giúp mình bài 2-3 với ạ

Lời giải:

Câu 2:

Áp dụng định lý $\cos$ ta được:

\(\begin{array}{l}
\quad BC^2 = SB^2 + SC^2 - 2SB.SC.\cos\widehat{BSC}\\
\Leftrightarrow BC^2 = 4a^2 + 9a^2 - 2.2a.3a.\cos60^\circ\\
\Leftrightarrow BC^2 = 7a^2\\
\Rightarrow BC = a\sqrt7
\end{array}\)

Ta có:

\(\begin{array}{l}
\quad \cos\alpha = \left|\cos\left(\overrightarrow{SA};\overrightarrow{BC}\right)\right|\\
\Leftrightarrow \cos\alpha = \dfrac{\left|\overrightarrow{SA}.\overrightarrow{BC}\right|}{SA.BC}\\
\Leftrightarrow \cos\alpha =\dfrac{\left|\overrightarrow{SA}.\left(\overrightarrow{SC}-\overrightarrow{SB}\right)\right|}{SA.BC}\\
\Leftrightarrow \cos\alpha =\dfrac{\left|\overrightarrow{SA}.\overrightarrow{SC}-\overrightarrow{SA}.\overrightarrow{SB}\right|}{SA.BC}\\
\Leftrightarrow \cos\alpha =\dfrac{a.3a.\cos90^\circ - a.2a.\cos60^\circ}{a.a\sqrt7}\\
\Leftrightarrow \cos\alpha = \dfrac{\sqrt7}{7}\\
\end{array}\)

Câu 3:

\(\begin{array}{l}
a)\quad \lim\limits_{x\to 0}\dfrac{\sqrt{3x+1} - 1}{x}\\
= \lim\limits_{x\to 0}\dfrac{\left(\sqrt{3x+1} - 1\right)\left(\sqrt{3x+1} + 1\right)}{x\left(\sqrt{3x+1} + 1\right)}\\
= \lim\limits_{x\to 0}\dfrac{3}{\sqrt{3x+1} + 1}\\
= \dfrac{3}{\sqrt{3.0+1} + 1}\\
= \dfrac32\\
\Rightarrow \begin{cases}a = 3\\b = 2\end{cases}\\
\Rightarrow P^2 = a^2 + b^2 = 13\\
b)\quad \dfrac{1}{x} + \dfrac{1}{x-a} + \dfrac{1}{x+b} = 0\qquad (*)\\
\Leftrightarrow \dfrac{(x-a)(x+b) + x(x+b) + x(x-a)}{x(x-a)(x+b)} = 0\\
\Leftrightarrow \dfrac{3x^2 - 2(a-b)x - ab}{x(x-a)(x+b)} = 0\\
\Leftrightarrow 3x^2 - 2(a-b)x - ab = 0\quad (**)\\
\text{Ta có:}\\
\quad \Delta_{(**)}' = (a-b)^2 + 3ab\\
\Leftrightarrow \Delta_{(**)}' = a^2 + ab + b^2\\
\Leftrightarrow \Delta_{(**)}' = \left(a + \dfrac b2\right)^2 + \dfrac{3b^2}{4} >0\quad \forall a, b >0\\
\Rightarrow (**)\ \text{luôn có hai nghiệm phân biệt}\\
\Rightarrow (*)\ \text{luôn có hai nghiệm phân biệt trên $(a;b)$}
\end{array}\)