ПОМОГИТЕ С 4 ЗАДАНИЕМ

№4 из 1-го варианта
Решаю по действиям.

1) $\sqrt{11-6 \sqrt{2} }= \sqrt{9-6 \sqrt{2} +2} = \sqrt{3^2-2*3* \sqrt{2} +( \sqrt{2} )^2} =$ $\sqrt{(3- \sqrt{2} )^2} =3- \sqrt{2}$

2) (6+2√2)·(3-√2) = 2·(3+√2)·(3-√2) = 2·(3²-√2²) = 2 · (9 - 2) = 2 · 7 = 14

3) $4,5- \sqrt{14} = \frac{9}{2}- \sqrt{14} = \frac{9-2 \sqrt{14} }{2} = \frac{7-2 \sqrt{7} \sqrt{2} +2}{2} = \frac{ \sqrt{7} ^2-2 \sqrt{7} * \sqrt{2} + \sqrt{2} ^2}{2} =$ $\frac{( \sqrt{7} - \sqrt{2} )^2}{2}$

4) $\frac{( \sqrt{7} - \sqrt{2} )^2}{2*( \sqrt{7} + \sqrt{2} )^{-2}} = \frac{( \sqrt{7} - \sqrt{2} )^2*( \sqrt{7} + \sqrt{2} )^2}{2} = \frac{( \sqrt{7} ^2- \sqrt{2} ^2)^2}{2} = \frac{(7-2)^2}{2} = \frac{25}{2} =12,5$

5) 14 - 12,5 = 1,5
Ответ: 1,5

Во втором варианте 4.

1) $\sqrt{7+4 \sqrt{3} } = \sqrt{4+4 \sqrt{3} +3} = \sqrt{2^2+2*2* \sqrt{3} + \sqrt{3} ^2} = \sqrt{(2+ \sqrt{3} )^2}$ $=2+ \sqrt{3}$

2) (4-2√3)·(2+√3) = 2·(2-√3)·(2+√3) = 2·(4-3) = 2 · 1 = 2

3) $\sqrt{7} -4= \frac{2 \sqrt{7} -8}{2}= \frac{-(1-2 \sqrt{7} +7)}{2} = \frac{-(1- \sqrt{7} )^2}{2}$

4) $\frac{-(1- \sqrt{7} )^2}{2*(1+ \sqrt{7} )^{-2}} = \frac{-(1- \sqrt{7} )^2*(1+ \sqrt{7} )^2}{2} = \frac{-(1-7)^2}{2}= \frac{-36}{2} =-18$

5) 2 - 18 = - 16