Упр.34 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1. Вычислить:
   
   1) $$\sqrt{9+\sqrt{17}}\cdot\sqrt{9-\sqrt{17}}$$;
   
   2) $$\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2$$;
   
   3) $$\left(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\right)^2$$;
   
   4) $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$.

1. $$\sqrt{9+\sqrt{17}}\cdot \sqrt{9-\sqrt{17}}=\sqrt{(9+\sqrt{17})(9-\sqrt{17})}$$
   $$=\sqrt{81-17}=\sqrt{64}=8.$$

2. $$\left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)^2$$
   $$=(3+\sqrt5)-2\sqrt{(3+\sqrt5)(3-\sqrt5)}+(3-\sqrt5)$$
   $$=6-2\sqrt{9-5}=6-2\sqrt4=6-2\cdot 2=2.$$

3. $$\left(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\right)^2$$
   $$=(5+\sqrt{21})+2\sqrt{(5+\sqrt{21})(5-\sqrt{21})}+(5-\sqrt{21})$$
   $$=10+2\sqrt{25-21}=10+2\sqrt4=10+2\cdot 2=14.$$

4. $$\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}-\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$$
   $$=\frac{(\sqrt3+\sqrt2)^2-(\sqrt3-\sqrt2)^2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}$$
   $$=\frac{(3+2\sqrt6+2)-(3-2\sqrt6+2)}{3-2}$$
   $$=\frac{4\sqrt6}{1}=4\sqrt6.$$

Ответ

1) $$8$$; 2) $$2$$; 3) $$14$$; 4) $$4\sqrt6$$.