Question

a) (√5 - √2)^2 + √40

б) √31b - √25b +3√b

в) (3√8+√18) * √2

г) (2a - √b) (2a + √b)

д) (√3+√2) (√3 - √2) - √24
СРОЧНООО!!!!!

Answer 1

Решение вашей задачи:

$(\sqrt{5}-\sqrt{2})^{2} + \sqrt{40} = \sqrt{5}^{2} -2\sqrt{5} \sqrt{2} +\sqrt{2}^{2}+\sqrt{40}= 5-2\sqrt{10}+2 +\sqrt{40}  = 5-2\sqrt{10}+2+2\sqrt{10} = 5+2 = 7$

$\sqrt{31b}-\sqrt{25b}+3\sqrt{b}=\sqrt{31b}-2\sqrt{b}$

$(3\sqrt{8}+\sqrt{18})\sqrt{2}=(6\sqrt{2}+\sqrt{18})\sqrt{2}=9\sqrt{2}\sqrt{2}=9*2=18$

$(2a-\sqrt{b})(2a+\sqrt{b}) = (2a)^{2}-\sqrt{b}^{2} =4a^{2}-\sqrt{b}^{2} = 4a^{2} -b$

$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})-\sqrt{24}=3-2-\sqrt{24}=1-2\sqrt{6}$