Упр.459 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1) 1/(v3-v2);
2) 2/(v11-v3);
3) (v7+v5)/(v7-v5);
4) (5-4v3)/(5v3-9).
Решить приведенное квадратное уравнение:
1) x^2-2v3 x-1=0;
2) x^2-2v5 x+1=0;
3) x^2+v2 x-4=0;
4) x^2-4v7 x+4=0.

1. $$\frac{1}{\sqrt{3}-\sqrt{2}}=\frac{1}{\sqrt{3}-\sqrt{2}}\cdot\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}.$$

   $$\frac{2}{\sqrt{11}-\sqrt{3}}=\frac{2}{\sqrt{11}-\sqrt{3}}\cdot\frac{\sqrt{11}+\sqrt{3}}{\sqrt{11}+\sqrt{3}}=\frac{2(\sqrt{11}+\sqrt{3})}{11-3}=\frac{\sqrt{11}+\sqrt{3}}{4}.$$

   $$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{(\sqrt{7}+\sqrt{5})^2}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}=\frac{7+2\sqrt{35}+5}{7-5}=\frac{12+2\sqrt{35}}{2}=6+\sqrt{35}.$$

   $$\frac{5-4\sqrt{3}}{5\sqrt{3}-9}=\frac{(5-4\sqrt{3})(5\sqrt{3}+9)}{(5\sqrt{3}-9)(5\sqrt{3}+9)}=\frac{25\sqrt{3}+45-60-36\sqrt{3}}{75-81}=\frac{-11\sqrt{3}-15}{-6}=\frac{11\sqrt{3}+15}{6}.$$

2. $$x^2-2\sqrt{3}\,x-1=0$$

   $$D=(2\sqrt{3})^2-4\cdot 1\cdot(-1)=12+4=16,$$

   $$x_{1,2}=\frac{2\sqrt{3}\pm 4}{2}=\sqrt{3}\pm 2.$$

   $$x_1=\sqrt{3}-2,\quad x_2=\sqrt{3}+2.$$

3. $$x^2-2\sqrt{5}\,x+1=0$$

   $$D=(2\sqrt{5})^2-4\cdot 1\cdot 1=20-4=16,$$

   $$x_{1,2}=\frac{2\sqrt{5}\pm 4}{2}=\sqrt{5}\pm 2.$$

   $$x_1=\sqrt{5}-2,\quad x_2=\sqrt{5}+2.$$

4. $$x^2+\sqrt{2}\,x-4=0$$

   $$D=(\sqrt{2})^2-4\cdot 1\cdot(-4)=2+16=18,$$

   $$\sqrt{D}=3\sqrt{2}.$$

   $$x_{1,2}=\frac{-\sqrt{2}\pm 3\sqrt{2}}{2}.$$

   $$x_1=\frac{-4\sqrt{2}}{2}=-2\sqrt{2},\quad x_2=\frac{2\sqrt{2}}{2}=\sqrt{2}.$$

5. $$x^2-4\sqrt{7}\,x+4=0$$

   $$D=(4\sqrt{7})^2-4\cdot 1\cdot 4=16\cdot 7-16=96,$$

   $$\sqrt{D}=4\sqrt{6}.$$

   $$x_{1,2}=\frac{4\sqrt{7}\pm 4\sqrt{6}}{2}=2\sqrt{7}\pm 2\sqrt{6}.$$

   $$x_1=2\sqrt{7}-2\sqrt{6},\quad x_2=2\sqrt{7}+2\sqrt{6}.$$

Ответ

1) $$\sqrt{3}+\sqrt{2};\ \frac{\sqrt{11}+\sqrt{3}}{4};\ 6+\sqrt{35};\ \frac{11\sqrt{3}+15}{6}.$$
2) $$x=\sqrt{3}\pm 2.$$
3) $$x=\sqrt{5}\pm 2.$$
4) $$x=-2\sqrt{2},\ \sqrt{2}.$$
5) $$x=2\sqrt{7}\pm 2\sqrt{6}.$$