Giả sử $T = f(2\sqrt{2}) - 2f(1)$; $MB = x$, $MA = 2x$, $MC = 3x$ với $0 < x < BC = \sqrt{2}$. Ta có $\cos BAM = \frac{1 + 4x^2 - x^2}{2 \cdot 1 \cdot 2x} = \frac{3x^2 + 1}{4x}$. \[ \cos MAC = \frac{1 + 4x^2 - 9x^2}{4x} = \frac{1 - 5x^2}{4x}. \] $C_6 = \frac{14\pi}{3}f(x) = e^x + 2x$. (1) \[ \Rightarrow \left(\frac{3x^2 + 1}{4x}\right)^2 + \left(\frac{1 - 5x^2}{4x}\right)^2 = 1 \Rightarrow 9x^4 + 6x^2 + 1 + 1 - 10x^2 + 25x^4 = 16. \] \[ \Rightarrow 34x^4 - 20x^2 + 2 = 0 \Leftrightarrow \left[ \begin{array}{l} x^2 = \frac{5 + 2\sqrt{2}}{17} > \frac{1}{5} \quad (1) \\ x^2 = \frac{5 - 2\sqrt{2}}{17} \end{array} \right]. \] $\Rightarrow \cos AMB = \frac{AM^2 + BM^2 - AB^2}{2AM \cdot BM} = \frac{4x^2 + x^2 - 1}{2 \cdot 2x \cdot x} = \frac{5x^2 - 1}{4x^2}$. \[ = \frac{5\left(\frac{5 - 2\sqrt{2}}{17}\right) - 1}{4\left(\frac{5 - 2\sqrt{2}}{17}\right)^2} = \frac{25 - 10\sqrt{2} - 1}{17} : \frac{20 - 8\sqrt{2} - \sqrt{2}}{17} = \frac{-\sqrt{2}}{2}. \] Vậy $AMB = 135^\circ$