*3.7. Решите уравнение a) \(\left(31x-\frac{1}{6}\right):\frac{5}{\sqrt{289}}=\frac{20}{5}:1,2\); б) \(\frac{\sqrt{3}+\sqrt{8}+\sqrt{3}-\sqrt{8}}{\sqrt{3}-\sqrt{8}+\sqrt{3}+\sqrt{8}}\cdot(x+\sqrt{2})=\frac{x-4\sqrt{2}}{\sqrt{8}}\); 6) \(\sqrt{15x}=\frac{\sqrt[3]{37}\cdot 75}{3\sqrt{15}}\); г) \(\frac{x}{\sqrt{4-2\sqrt{3}}}-\frac{x}{\sqrt{(1+\sqrt{3})^2}}=\left(\frac{7}{3}-\frac{3}{5}\right)\left(1\right)\left(\frac{1}{13}+0,5\right)\)

Решение:

а) \(\left(31x-\frac{1}{6}\right):\frac{5}{\sqrt{289}}=\frac{20}{5}:1,2\)

1. \( \sqrt{289}=17 \)
2. \( \frac{20}{5}=4 \)
3. \( 1,2 = \frac{12}{10} = \frac{6}{5} \)
4. Уравнение приобретает вид: \( \left(31x-\frac{1}{6}\right):\frac{5}{17}=4:\frac{6}{5} \)
5. \( \left(31x-\frac{1}{6}\right)\cdot\frac{17}{5} = 4\cdot\frac{5}{6} \)
6. \( \left(31x-\frac{1}{6}\right)\cdot\frac{17}{5} = \frac{20}{6} = \frac{10}{3} \)
7. \( 31x-\frac{1}{6} = \frac{10}{3}\cdot\frac{5}{17} \)
8. \( 31x-\frac{1}{6} = \frac{50}{51} \)
9. \( 31x = \frac{50}{51} + \frac{1}{6} = \frac{100+17}{102} = \frac{117}{102} \)
10. \( x = \frac{117}{102\cdot 31} = \frac{117}{3162} \)

б) \(\frac{\sqrt{3}+\sqrt{8}+\sqrt{3}-\sqrt{8}}{\sqrt{3}-\sqrt{8}+\sqrt{3}+\sqrt{8}}\cdot(x+\sqrt{2})=\frac{x-4\sqrt{2}}{\sqrt{8}}\ )

1. \( \sqrt{3}+\sqrt{8}+\sqrt{3}-\sqrt{8} = 2\sqrt{3} \)
2. \( \sqrt{3}-\sqrt{8}+\sqrt{3}+\sqrt{8} = 2\sqrt{3} \)
3. \( \sqrt{8} = 2\sqrt{2} \)
4. Уравнение приобретает вид: \( \frac{2\sqrt{3}}{2\sqrt{3}}\cdot(x+\sqrt{2})=\frac{x-4\sqrt{2}}{2\sqrt{2}} \)
5. \( 1\cdot(x+\sqrt{2}) = \frac{x-4\sqrt{2}}{2\sqrt{2}} \)
6. \( x+\sqrt{2} = \frac{x}{2\sqrt{2}} - \frac{4\sqrt{2}}{2\sqrt{2}} \)
7. \( x+\sqrt{2} = \frac{x}{2\sqrt{2}} - 2 \)
8. \( x - \frac{x}{2\sqrt{2}} = -2 - \sqrt{2} \)
9. \( x(1-\frac{1}{2\sqrt{2}}) = -(2+\sqrt{2}) \)
10. \( x(\frac{2\sqrt{2}-1}{2\sqrt{2}}) = -(2+\sqrt{2}) \)
11. \( x = \frac{-(2+\sqrt{2})2\sqrt{2}}{2\sqrt{2}-1} = \frac{-4\sqrt{2}-4}{2\sqrt{2}-1} \)
12. \( x = \frac{(-4\sqrt{2}-4)(2\sqrt{2}+1)}{(2\sqrt{2}-1)(2\sqrt{2}+1)} = \frac{-16-4\sqrt{2}-8\sqrt{2}-4}{8-1} = \frac{-20-12\sqrt{2}}{7} \)

6) \(\sqrt{15x}=\frac{\sqrt[3]{37}\cdot 75}{3\sqrt{15}}\ )

1. \( \sqrt{15x} = \frac{\sqrt[3]{37} \cdot 75}{3\sqrt{15}} \)
2. \( 15x = \left(\frac{\sqrt[3]{37} \cdot 75}{3\sqrt{15}}\right)^2 = \frac{(\sqrt[3]{37})^2 \cdot 75^2}{9 \cdot 15} \)
3. \( 15x = \frac{37^{2/3} \cdot 5625}{135} \)
4. \( x = \frac{37^{2/3} \cdot 5625}{135 \cdot 15} = \frac{37^{2/3} \cdot 5625}{2025} = 37^{2/3} \cdot \frac{5625}{2025} = 37^{2/3} \cdot \frac{25}{9} \)

г) \(\frac{x}{\sqrt{4-2\sqrt{3}}}-\frac{x}{\sqrt{(1+\sqrt{3})^2}}=\left(\frac{7}{3}-\frac{3}{5}\right)\left(1\right)\left(\frac{1}{13}+0,5\right)\)

1. \( \sqrt{4-2\sqrt{3}} = \sqrt{(\sqrt{3}-1)^2} = \sqrt{3}-1 \)
2. \( \sqrt{(1+\sqrt{3})^2} = 1+\sqrt{3} \)
3. \( \frac{7}{3}-\frac{3}{5} = \frac{35-9}{15} = \frac{26}{15} \)
4. \( \frac{1}{13}+0,5 = \frac{1}{13}+\frac{1}{2} = \frac{2+13}{26} = \frac{15}{26} \)
5. Уравнение приобретает вид: \( \frac{x}{\sqrt{3}-1} - \frac{x}{1+\sqrt{3}} = \frac{26}{15} \cdot 1 \cdot \frac{15}{26} \)
6. \( \frac{x(\sqrt{3}+1)-x(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = 1 \)
7. \( \frac{x\sqrt{3}+x-x\sqrt{3}+x}{3-1} = 1 \)
8. \( \frac{2x}{2} = 1 \)
9. \( x = 1 \)

Ответ: а) \( x = \frac{117}{3162} \); б) \( x = \frac{-20-12\sqrt{2}}{7} \); 6) \( x = 37^{2/3} \cdot \frac{25}{9} \); г) x = 1.