Question

помогите с алгеброй​

Answer 1

Ответ:

1)

$\sqrt{26} > 5 \\ 6\sqrt{3} > 5\sqrt{4}$

2)

а) $\sqrt{121} - 10\sqrt{6,4} * \sqrt{0.1} = 11 - 10\sqrt{0,64} = 11 - 10 * 0,8 = 11 - 8 = 3$

б) $2\sqrt{5} - \sqrt{45} +\sqrt{80} = 2\sqrt{5} - 3\sqrt{5} + 4\sqrt{5} = 3\sqrt{5}$

3)

$(\sqrt{10} -8)(\sqrt{10} +8) = 10 -64 = -54 \\(3\sqrt{6} + \sqrt{12} )\sqrt{3} = 9\sqrt{2} + \sqrt{36} = 9\sqrt{2} + 6 \\(5-\sqrt{2})^{2} = 25 -10\sqrt{2} + 2 = 27 - 10\sqrt{2}$

4)

$\frac{a-3\sqrt{a} }{a-9} = \frac{\sqrt{a}( \sqrt{a}-3) }{(\sqrt{a}-3)(\sqrt{a} +3) } =\\= \frac{\sqrt{a} }{\sqrt{a} + 3}$

5)

$\frac{4}{2\sqrt{3} + 1} - \frac{4}{2\sqrt{3}-1 } = \\=\frac{8\sqrt{3}-4 - 8\sqrt{3} - 4 }{11} = \frac{-8}{11}$