Question

Помогайте1 Освободитесь от иррациональности в знаменателе Б) 4/√7-√3 И ещё Упрастите выражение : А) 7√3-√108+√75 Б) (√а - √b) - a

Answer 1

$\frac{4( \sqrt{7} + \sqrt{3} )}{4} = \sqrt{7} + \sqrt{3}$
$7 \sqrt{3} - 6 \sqrt{3} + 5 \sqrt{3} = 6 \sqrt{3}$
$\sqrt{a} - \sqrt{b} - a$

Answer 2

I can help you with some math problems.

To get rid of the irrationality in the denominator of B) $\frac{4}{\sqrt{7}-\sqrt{3}}$, you need to multiply the numerator and denominator by the conjugate binomial of the denominator, which is $\sqrt{7}+\sqrt{3}$. This will give you:

$$\frac{4}{\sqrt{7}-\sqrt{3}} \cdot \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} = \frac{4(\sqrt{7}+\sqrt{3})}{7-3} = \frac{2(\sqrt{7}+\sqrt{3})}{2} = \sqrt{7}+\sqrt{3}$$

To simplify the expression A) $7\sqrt{3}-\sqrt{108}\sqrt{75}$, you need to factor out the perfect squares from the radicals and simplify them. This will give you:

$$7\sqrt{3}-\sqrt{108}\sqrt{75} = 7\sqrt{3}-\sqrt{36 \cdot 3}\sqrt{25 \cdot 3} = 7\sqrt{3}-6\sqrt{3} \cdot 5\sqrt{3} = 7\sqrt{3}-90\sqrt{3} = -83\sqrt{3}$$

To simplify the expression B) $(\sqrt{a}-\sqrt{b})-a$, you need to expand the parentheses and combine the like terms. This will give you:

$$(\sqrt{a}-\sqrt{b})-a = \sqrt{a}-\sqrt{b}-a$$

There is nothing more you can do to simplify this expression, unless you know the values of $a$ and $b$.

I hope this helps you with your math homework. If you need more help, you can check out some online resources, such as [this one](https://algeomath.ru/%d0%bc%d0%b5%d1%80%d0%b7%d0%bb%d1%8f%d0%ba-8-%d0%ba%d0%bb%d0%b0%d1%81%d1%81-%d0%ba%d0%be%d0%bd%d1%82%d1%80%d0%be%d0%bb%d1%8c%d0%bd%d0%b0%d1%8f-4/) or [this one](https://ru.wikihow.com/%D0%B8%D0%B7%D0%B1%D0%B0%D0%B2%D0%B8%D1%82%D1%8C%D1%81%D1%8F-%D0%BE%D1%82-%D0%B8%D1%80%D1%80%D0%B0%D1%86%D0%B8%D0%BE%D0%BD%D0%B0%D0%BB%D1%8C%D0%BD%D0%BE%D1%81%D1%82%D0%B8-%D0%B2-%D0%B7%D0%BD%D0%B0%D0%BC%D0%B5%D0%BD%D0%B0%D1%82%D0%B5%D0%BB%D0%B5).