So sánh p&i club và công ty bảo hiểm năm 2024

\(\begin{array}{l}P = a - \left\{ {\left( {a - 3} \right) - \left[ {\left( {a + 3} \right) - \left( { - a - 2} \right)} \right]} \right\}\\\,\,\,\, = a - \left[ {\left( {a - 3} \right) - \left( {a + 3 + a + 2} \right)} \right]\\\,\,\,\, = a - \left( {a - 3 - a - 3 - a - 2} \right)\\\,\,\,\, = a - \left( { - a - 8} \right)\\\,\,\,\, = a + a + 8\\\,\,\,\, = 2a + 8\end{array}\)

\(\begin{array}{l}Q = \left[ {a + \left( {a + 3} \right)} \right] + \left[ {\left( {a + 2} \right) - \left( {a - 2} \right)} \right]\\\,\,\,\,\, = \left( {a + a + 3} \right) + \left( {a + 2 - a + 2} \right)\\\,\,\,\, = \left( {2a + 3} \right) + 4\\\,\,\,\, = 2a + 3 + 4\\\,\,\,\, = 2a + 7\end{array}\)

\(p=a-\left\{\left(a-3\right)-\left[\left(a+3\right)-\left(-a-2\right)\right]\right\}\)

\(=a-\left\{a-3-\left[a+3+a+2\right]\right\}\)

\(=a-\left\{a-3-a-3-a-2\right\}\)

\(=a-\left\{-a-8\right\}\)

\(=a+a+8\)

\(=2a+8\)

\(Q=\left[a+\left(a+3\right)\right]-\left[\left(a+2\right)-\left(a-2\right)\right]\)

\(=\left[a+a+3\right]-\left[a+2-a+2\right]\)

\(=2a+3-4\)

\(=2a-1\)

Xét hiệu \(P-Q=\left(2a+8\right)-\left(2a-1\right)\)

\(=2a+8-2a+1\)

\(=9>0\)

Vậy: \(P>Q\)

Xét \(P - 2 = \dfrac{{2\sqrt x }}{{\sqrt x + 1}} - 2 = \dfrac{{2\sqrt x - 2\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\)

\(P - 2 = \dfrac{{2\sqrt x - 2\sqrt x - 2}}{{\sqrt x + 1}} = \dfrac{{ - 2}}{{\sqrt x + 1}}\)

Ta có \(x \ge 0 \Rightarrow \sqrt x \ge 0 \Rightarrow \sqrt x + 1 \ge 1 \Rightarrow \sqrt x + 1 > 0\)

\( \Rightarrow \dfrac{{ - 2}}{{\sqrt x + 1}} < 0 \Leftrightarrow P - 2 < 0 \Leftrightarrow P < 2\,\,\forall x\,\,tm\,\left\{ \begin{array}{l}x \ge 0\\x \ne 1\end{array} \right.\).