Упр.34 ГДЗ Колягин Ткачёва 9 класс (Алгебра)
1) v(9+v17)·v(9-v17); 2) (v(3+v5)-v(3-v5))^2;
3) (v(5+v21)+v(5-v21)^2; 4) (v3+v2)/(v3-v2)-(v3-v2)/(v3+v2).
$$\sqrt{9+\sqrt{17}}\cdot \sqrt{9-\sqrt{17}}=\sqrt{(9+\sqrt{17})(9-\sqrt{17})}$$
$$=\sqrt{81-17}=\sqrt{64}=8.$$$$\left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)^2$$
$$=(3+\sqrt5)-2\sqrt{(3+\sqrt5)(3-\sqrt5)}+(3-\sqrt5)$$
$$=6-2\sqrt{9-5}=6-2\sqrt4=6-2\cdot 2=2.$$$$\left(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\right)^2$$
$$=(5+\sqrt{21})+2\sqrt{(5+\sqrt{21})(5-\sqrt{21})}+(5-\sqrt{21})$$
$$=10+2\sqrt{25-21}=10+2\sqrt4=10+2\cdot 2=14.$$$$\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}-\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$$
$$=\frac{(\sqrt3+\sqrt2)^2-(\sqrt3-\sqrt2)^2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}$$
$$=\frac{(3+2\sqrt6+2)-(3-2\sqrt6+2)}{3-2}$$
$$=\frac{4\sqrt6}{1}=4\sqrt6.$$
Ответ
1) $$8$$; 2) $$2$$; 3) $$14$$; 4) $$4\sqrt6$$.