Вычислить : 3/7(4-√2)(3/(1-√2)+ (2/(2 +√2) +3/(3-2√2))

Решите задачу:

$\frac{3(4 - \sqrt{2} )}{7} ( \frac{3}{1 - \sqrt{2} } + \frac{2}{2 + \sqrt{2} } + \frac{3}{3 - 2 \sqrt{2} } ) = \frac{3(4 - \sqrt{2}) }{7} ( \frac{3(1 + \sqrt{2} )}{(1 - \sqrt{2} )(1 + \sqrt{2}) } + \frac{2(2 - \sqrt{2} )}{(2 + \sqrt{2} )(2 - \sqrt{2} )} + \frac{3(3 + 2 \sqrt{2}) }{(3 - 2 \sqrt{2} )(3 + 2 \sqrt{2}) } ) = \frac{3(4 - \sqrt{2} )}{7} ( \frac{3(1 + \sqrt{2}) }{1 - 2} + \frac{2(2 - \sqrt{2} )}{4 - 2} + \frac{3(3 + 2 \sqrt{2}) }{9 - 8} = \frac{3(4 - \sqrt{2} )}{7} ( - 3 - 3 \sqrt{2} + 2 - \sqrt{2} + 9 + 6 \sqrt{2} ) = \frac{3(4 - \sqrt{2}) }{7} (2 \sqrt{2} + 8) = \frac{3}{7} \times 2(4 - \sqrt{2} )(4 + \sqrt{2} ) = \frac{6}{7} (16 - 2) = \frac{6}{7} \times 14 = 6 \times 2 = 12$