Предмет: Алгебра, автор: davletshinalayy

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Приложения:

Ответы

Автор ответа: axatar

Ответ:

1. Вычислите.

а)

$\displaystyle \tt 6 \cdot \sqrt{1,21} -2 \cdot (\sqrt{2})^2 = 6 \cdot \sqrt{(1,1)^2} -2 \cdot 2 = 6 \cdot 1,1 -4 = 6,6-4 = 2,6;$

б)

$\displaystyle \tt 8 \cdot \sqrt{2\frac{1}{4} } -3 \cdot \sqrt{5\frac{4}{9} } = 8 \cdot \sqrt{\frac{9}{4} } -3 \cdot \sqrt{\frac{49}{9} }= \\\\=8 \cdot \sqrt{\bigg (\frac{3}{2} \bigg)^2} -3 \cdot \sqrt{\bigg (\frac{7}{3} \bigg )^2 }=8 \cdot \frac{3}{2} - 3 \cdot \frac{7}{3}= 12-7=5;$

в)

$\displaystyle \tt (\sqrt{18}-\sqrt{2})^2 = (\sqrt{9 \cdot 2}-\sqrt{2})^2 = (\sqrt{3^2 \cdot 2}-\sqrt{2})^2 = (3 \cdot \sqrt{2}-\sqrt{2})^2=\\\\=(2 \cdot \sqrt{2})^2 = 4 \cdot 2 = 8.$

2. Сравнить числа.

а)

$\displaystyle \tt \sqrt{6}$ и $\displaystyle \tt \sqrt{5}$ ⇔ $\displaystyle \tt (\sqrt{6})^2$ и $\displaystyle \tt (\sqrt{5})^2$ ⇔ 6 и 5 ⇔ 6>5 ⇒

б)

$\displaystyle \tt \sqrt{1,5}$ и $\displaystyle \tt \sqrt{1\frac{2}{3} }$ ⇔ $\displaystyle \tt \bigg (\sqrt{\frac{3}{2} } \bigg )^2$ и $\displaystyle \tt \bigg (\sqrt{\frac{5}{3}} \bigg )^2$ ⇔ $\displaystyle \tt \frac{3}{2}$ и $\displaystyle \tt \frac{5}{3}$ ⇔ $\displaystyle \tt \frac{9}{6}$ < $\displaystyle \tt \frac{10}{6}$ ⇒

3. Упростите.

а)

$\displaystyle \tt 3 \cdot \sqrt{2} +\sqrt{50} -\sqrt{18} = 3 \cdot \sqrt{2} +\sqrt{25 \cdot 2} -\sqrt{9 \cdot 2} =3 \cdot \sqrt{2} +\sqrt{5^2 \cdot 2} -\sqrt{3^2 \cdot 2} =\\\\= 3 \cdot \sqrt{2} +5 \cdot \sqrt{2} -3 \cdot \sqrt{2} = 5 \cdot \sqrt{2};$

б)

$\displaystyle \tt (2 \cdot \sqrt{5} -\sqrt{27}) \cdot \sqrt{3} -2 \cdot \sqrt{15}=2 \cdot \sqrt{5} \cdot \sqrt{3} -\sqrt{27} \cdot \sqrt{3} -2\cdot \sqrt{15}=\\\\=2 \cdot \sqrt{15} -2\cdot \sqrt{15}-\sqrt{27 \cdot 3}=-\sqrt{81}=-\sqrt{9^2}= -9.$

4. Сократите дробь.

а)

$\displaystyle \tt \frac{\sqrt{7} -2}{\sqrt{14} -2 \cdot \sqrt{2} }=\frac{\sqrt{7} -2}{\sqrt{7 \cdot 2} -2 \cdot \sqrt{2} }=\frac{\sqrt{7} -2}{(\sqrt{7} -2) \cdot \sqrt{2} }=\frac{1}{\sqrt{2} }=\frac{\sqrt{2}}{2 };$

б)

$\displaystyle \tt \frac{3+\sqrt{3}}{\sqrt{15} + \sqrt{5} }=\frac{3+\sqrt{3}}{\sqrt{3 \cdot 5} + \sqrt{5} }=\frac{(\sqrt{3}+1) \cdot \sqrt{3}}{(\sqrt{3 } + 1)\cdot \sqrt{5} }=\frac{\sqrt{3} }{\sqrt{5} } =\frac{\sqrt{15} }{5 } ;$

в)

$\displaystyle \tt \frac{x^2-3}{\sqrt{3} \cdot x+ 3 }=\frac{(x-\sqrt{3}) \cdot (x+\sqrt{3})}{\sqrt{3} \cdot (x+ \sqrt{3} )}=\frac{(x-\sqrt{3}) }{\sqrt{3} }=\frac{\sqrt{3} \cdot(x-\sqrt{3}) }{3}=\frac{\sqrt{3} \cdot x-3 }{3}.$

5*. Освободите от иррациональности в знаменателе дроби.

а)

$\displaystyle \tt \frac{3}{\sqrt{6}} = \frac{3 \cdot \sqrt{6} }{\sqrt{6} \cdot \sqrt{6} } =\frac{3 \cdot \sqrt{6} }{6} =\frac{\sqrt{6} }{2};$

б)

$\displaystyle \tt \frac{1}{\sqrt{7}-\sqrt{5}} = \frac{1 \cdot (\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5}) \cdot (\sqrt{7}+\sqrt{5})} = \frac{\sqrt{7}+\sqrt{5}}{7-5} = \frac{\sqrt{7}+\sqrt{5}}{2};$

в)

$\displaystyle \tt \frac{1}{\sqrt{4+2\cdot \sqrt{3}}}=\frac{1 \cdot \sqrt{4-2 \cdot \sqrt{3}}}{\sqrt{4+2 \cdot \sqrt{3}} \cdot \sqrt{4-2 \cdot \sqrt{3}}} = \frac{\sqrt{4-2 \cdot \sqrt{3}}}{\sqrt{16-4 \cdot 3}} =\\\\\\=\frac{\sqrt{4-2 \cdot \sqrt{3}}}{\sqrt{16-12}}=\frac{\sqrt{4-2 \cdot \sqrt{3}}}{\sqrt{4}} =\frac{\sqrt{4-2 \cdot \sqrt{3}}}{\sqrt{2^2}} =\frac{\sqrt{4-2 \cdot \sqrt{3}}}{2}=\\\\=\frac{\sqrt{3-2 \cdot \sqrt{3}+1}}{2}=\frac{\sqrt{(\sqrt{3}-1)^2}}{2}=\frac{|\sqrt{3}-1|}{2}=\frac{\sqrt{3}-1}{2}.$