Cho hình chóp \(S.ABC\) có đáy \(ABC\) là tam giác vuông tại \(B\), \(AB = 3a\), \(BC = 4a\) và \(SA \bot \left( {ABC} \right)\). Góc giữa đường thẳng \(SC\) và mặt phẳng \(\left( {ABC} \right)\) bằng \(60^\circ \). Gọi \(M\) là trung điểm của cạnh \(AC\). Khoảng cách giữa hai đường thẳng \(AB\) và \(SM\) bằng
A. \(\frac{{10\sqrt 3 a}}{{\sqrt {79} }}\).                
B. \(\frac{{5a}}{2}\).          
C. \(5\sqrt 3 a\).                 
D. \(\frac{{5\sqrt 3 a}}{{\sqrt {79} }}\).

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Cho biểu thức: \(P = \frac{{2{x^2} - 1}}{{{x^2} + x}} - \frac{{x - 1}}{x} + \frac{3}{{x + 1}}\)
1. Rút gọn \(P\) .
2. Tìm x để \(P = 0\)
3. Tính giá trị biểu thức \(P\) khi \(x\) thỏa mãn: \({x^2} - x = 0\).
4. Tìm giá trị lớn nhất của biểu thức \(Q = \frac{1}{{{x^2} - 9}}.P\)
A.\(\begin{array}{l}1)\,\,P = \frac{{x + 3}}{{x + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = 2\\2)\,\,x =  - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{{ - 1}}{4}\end{array}\)
B.\(\begin{array}{l}1)\,\,P = \frac{{x + 3}}{{x + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = 2\\2)\,\,x = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{1}{4}\end{array}\)
C.\(\begin{array}{l}1)\,\,P = \frac{{x - 3}}{{x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P =  - 2\\2)\,\,x =  - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{{ - 1}}{4}\end{array}\)
D.\(\begin{array}{l}1)\,\,P = \frac{{x - 3}}{{x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P =  - 2\\2)\,\,x =  - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{1}{4}\end{array}\)