Задание 22, стр 27

Question

Значения каких выражений является иррациональным числом?
$$a=\sqrt{5}-\frac{1}{\sqrt{5}-2}$$
$$b=\sqrt{3}-\frac{1}{2+\sqrt{3}}$$
$$c=\sqrt{5}+\frac{1}{\sqrt{5}+2}$$

Gimli · Accepted Answer

$$a=\sqrt{5}-\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}(\sqrt{5}-2)}{\sqrt{5}-2}-\frac{1}{\sqrt{5}-2}=\frac{5-2 \sqrt{5}-1}{\sqrt{5}-2}= \frac{4-2 \sqrt{5}}{\sqrt{5}-2}=\frac{2(2-\sqrt{5})}{\sqrt{5}-2}=\frac{-2(\sqrt{5}-2)}{\sqrt{5}-2}=$$
$$=-2 
otin \mathbb{R} $$
$$b=\sqrt{3}-\frac{1}{2+\sqrt{3}}=\frac{\sqrt{3}(2+\sqrt{3})}{2+\sqrt{3}}-\frac{1}{2+\sqrt{3}}=\frac{2 \sqrt{3}+3-1}{2+\sqrt{3}} = \frac{2\sqrt{3}+2}{2+\sqrt{3}} = \frac{(2\sqrt{3}+2)(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=$$
$$=\frac{4\sqrt{3}-6+4-2\sqrt{3}}{4-3} = 2\sqrt{3}-2\in \mathbb{R}$$
$$c=\sqrt{5}+\frac{1}{\sqrt{5}+2}=\frac{\sqrt{5}(\sqrt{5}+2)}{\sqrt{5}+2}-\frac{1}{\sqrt{5}+2}=\frac{5 \sqrt{5}+10+1}{\sqrt{5}+2} = \frac{5\sqrt{5}+11}{\sqrt{5}+2} = \frac{(5\sqrt{5}+11)(\sqrt{5}-2)}{5-4}=$$
% = \frac{(2\sqrt{3}+2)(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=$$
$$=\frac{5\cdot 5 + 11\sqrt{5} - 10\sqrt{5} -22}{1} =5\cdot 5 + 11\sqrt{5} - 10\sqrt{5} -22 = 25+\sqrt{5} -22 = 3+\sqrt{5}
otin \mathbb{R}$$
	extbf{Ответ: }$a; c.$

Задание 22, стр 27

Answers

Comments

Задание 22, стр 27

Answers

Comments

Add answer