ПОМОГИТЕ!!!АЛГЕБРА 8 КЛАСС!!!СРОЧНО!!!

1.
а)
$6 \sqrt{1 \times \frac{7}{9} } - 4 = 6 \sqrt{ \frac{16}{9} } - 4 = 6 \times \frac{4}{3} - 4 = 2 \times 4 - 4 = 8 - 4 = 4$
б)
$\sqrt{7.2} \times \sqrt{20} = \sqrt{0.36 \times 20} \times \sqrt{20} = 0.6 \times 20 = 12$
в)
$\frac{ \sqrt{216} }{ \sqrt{6} } = \frac{ \sqrt{6} \sqrt{36} }{ \sqrt{6} } = \sqrt{36} = 6$
г)
$\sqrt{ {5}^{4} \times {3}^{2} } = {5}^{2} \times 3 = 25 \times 3 = 75$
2.
а)
$4 \sqrt{20} - \sqrt{125} = 4 \sqrt{4 \times 5} - \sqrt{25 \times 5} = 8 \sqrt{5} - 5 \sqrt{5} = 3 \sqrt{5}$
б)
$(3 \sqrt{6} + \sqrt{12} ) \sqrt{3} = (3 \sqrt{6} + \sqrt{4 \times 3}) \sqrt{3} = (3 \sqrt{6} + 2 \sqrt{3} ) \sqrt{3} = 3 \sqrt{18} + 6 = 3 \sqrt{9 \times 2} + 6 = 9 \sqrt{2} + 6$
в)
${(5 - \sqrt{2)} }^{2} = {5}^{2} - 2 \times 5 \sqrt{2} + { \sqrt{2} }^{2} = 25 - 10 \sqrt{2} + 2 = 27 - 10 \sqrt{2}$
3.
а)
$12 \sqrt{3} = \sqrt{144 \times 3} = \sqrt{432}$
б)
$- 9 \sqrt{2} = - \sqrt{81 \times 2} = - \sqrt{162}$
4.
$\sqrt{ {x}^{2} - 6x + 9 } = \sqrt{ {(x - 3)}^{2} } = |x - 3|$
|2.6 - 3| = |-0.4| = 0.4

Дальше сама. :)