Question

Пожалуйста помогите срочно надо

Answer 1

Ответ:

Объяснение:

1)

$\dfrac{a-\sqrt{3} }{a^{2} -3} =\dfrac{a-\sqrt{3}}{a^{2} -(\sqrt{3})^{2} } =\dfrac{a -\sqrt{3} }{(a -\sqrt{3})(a +\sqrt{3}) }=\dfrac{1}{a +\sqrt{3}}$

2)

$\dfrac{b+\sqrt{7} }{b^{2} -7} =\dfrac{b+\sqrt{7}}{b^{2} -(\sqrt{7})^{2} } =\dfrac{b +\sqrt{7} }{(b-\sqrt{7})(b +\sqrt{7}) }=\dfrac{1}{b-\sqrt{7}}$

3)

$\dfrac{a+\sqrt{3a} }{a -3} =\dfrac{(\sqrt{a})^{2} +\sqrt{a} \: \cdot\sqrt{3}}{(\sqrt{a} )^{2} -(\sqrt{3})^{2} } =\dfrac{\sqrt{a} (\sqrt{a}+ \sqrt{3} )}{(\sqrt{a} -\sqrt{3})(\sqrt{a} +\sqrt{3}) }=\dfrac{\sqrt{a} }{\sqrt{a} -\sqrt{3}}$

4)

$\dfrac{a^{2} -a\sqrt{5} }{a^{2}-5}= \dfrac{a^{2} -a\sqrt{5} }{a^{2}-(\sqrt{5})^{2} }=\dfrac{a\left(a-\sqrt{5}\right)}{\left(a+\sqrt{5}\right)\left(a-\sqrt{5}\right)}=\dfrac{a}{a+\sqrt{5}}$

5)

$\dfrac{3\sqrt{x} -2\sqrt{a}}{9x-4a} =\dfrac{\sqrt{9x} -\sqrt{4a} }{(\sqrt{9x})^{2} -(\sqrt{4a})^{2} } =\dfrac{\sqrt{9x} -\sqrt{4a}}{(\sqrt{9x} -\sqrt{4a})(\sqrt{9x} +\sqrt{4a})} =\dfrac{1}{\sqrt{9x} +\sqrt{4a}}$

6)

$\dfrac{7+\sqrt{3c} }{49-3c} =\dfrac{7+\sqrt{3c} }{7^{2} -(\sqrt{3c} )^{2} } =\dfrac{7+\sqrt{3c}}{(7-\sqrt{3c})(7+\sqrt{3c})} =\dfrac{1}{7-\sqrt{3c}}$