メンタ 15 ~395 21- x=ath N394 1 20-X-X N396 3+=x N397 N398 L+X X 313 + XX+2 V399 N400 文+x+2 1+2=7 3 3 =1 h+x X

Решаю уравнения.

1. № 394

   $$\frac{x}{20 - x} = \frac{1}{x}$$

   Решение:

   $$x^2 = 20 - x$$

   $$x^2 + x - 20 = 0$$

   $$D = 1^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81$$

   $$x_1 = \frac{-1 + \sqrt{81}}{2 \cdot 1} = \frac{-1 + 9}{2} = \frac{8}{2} = 4$$

   $$x_2 = \frac{-1 - \sqrt{81}}{2 \cdot 1} = \frac{-1 - 9}{2} = \frac{-10}{2} = -5$$

   Ответ: x = 4, x = -5

2. № 395

   $$4 + \frac{21}{x} = x$$

   Решение:

   $$4x + 21 = x^2$$

   $$x^2 - 4x - 21 = 0$$

   $$D = (-4)^2 - 4 \cdot 1 \cdot (-21) = 16 + 84 = 100$$

   $$x_1 = \frac{4 + \sqrt{100}}{2 \cdot 1} = \frac{4 + 10}{2} = \frac{14}{2} = 7$$

   $$x_2 = \frac{4 - \sqrt{100}}{2 \cdot 1} = \frac{4 - 10}{2} = \frac{-6}{2} = -3$$

   Ответ: x = 7, x = -3

3. № 396

   $$3 + \frac{10}{x} = x$$

   Решение:

   $$3x + 10 = x^2$$

   $$x^2 - 3x - 10 = 0$$

   $$D = (-3)^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49$$

   $$x_1 = \frac{3 + \sqrt{49}}{2 \cdot 1} = \frac{3 + 7}{2} = \frac{10}{2} = 5$$

   $$x_2 = \frac{3 - \sqrt{49}}{2 \cdot 1} = \frac{3 - 7}{2} = \frac{-4}{2} = -2$$

   Ответ: x = 5, x = -2

4. № 397

   $$\frac{6}{x} + \frac{6}{x+1} = 5$$

   Решение:

   $$\frac{6(x+1) + 6x}{x(x+1)} = 5$$

   $$\frac{6x + 6 + 6x}{x^2 + x} = 5$$

   $$\frac{12x + 6}{x^2 + x} = 5$$

   $$12x + 6 = 5(x^2 + x)$$

   $$12x + 6 = 5x^2 + 5x$$

   $$5x^2 + 5x - 12x - 6 = 0$$

   $$5x^2 - 7x - 6 = 0$$

   $$D = (-7)^2 - 4 \cdot 5 \cdot (-6) = 49 + 120 = 169$$

   $$x_1 = \frac{7 + \sqrt{169}}{2 \cdot 5} = \frac{7 + 13}{10} = \frac{20}{10} = 2$$

   $$x_2 = \frac{7 - \sqrt{169}}{2 \cdot 5} = \frac{7 - 13}{10} = \frac{-6}{10} = -0.6$$

   Ответ: x = 2, x = -0.6

5. № 398

   $$\frac{3}{x} + \frac{3}{x+2} = 4$$

   Решение:

   $$\frac{3(x+2) + 3x}{x(x+2)} = 4$$

   $$\frac{3x + 6 + 3x}{x^2 + 2x} = 4$$

   $$\frac{6x + 6}{x^2 + 2x} = 4$$

   $$6x + 6 = 4(x^2 + 2x)$$

   $$6x + 6 = 4x^2 + 8x$$

   $$4x^2 + 8x - 6x - 6 = 0$$

   $$4x^2 + 2x - 6 = 0$$

   $$2x^2 + x - 3 = 0$$

   $$D = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25$$

   $$x_1 = \frac{-1 + \sqrt{25}}{2 \cdot 2} = \frac{-1 + 5}{4} = \frac{4}{4} = 1$$

   $$x_2 = \frac{-1 - \sqrt{25}}{2 \cdot 2} = \frac{-1 - 5}{4} = \frac{-6}{4} = -1.5$$

   Ответ: x = 1, x = -1.5

6. № 399

   $$\frac{1}{x} + \frac{2}{x+2} = 1$$

   Решение:

   $$\frac{1(x+2) + 2x}{x(x+2)} = 1$$

   $$\frac{x + 2 + 2x}{x^2 + 2x} = 1$$

   $$\frac{3x + 2}{x^2 + 2x} = 1$$

   $$3x + 2 = x^2 + 2x$$

   $$x^2 + 2x - 3x - 2 = 0$$

   $$x^2 - x - 2 = 0$$

   $$D = (-1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9$$

   $$x_1 = \frac{1 + \sqrt{9}}{2 \cdot 1} = \frac{1 + 3}{2} = \frac{4}{2} = 2$$

   $$x_2 = \frac{1 - \sqrt{9}}{2 \cdot 1} = \frac{1 - 3}{2} = \frac{-2}{2} = -1$$

   Ответ: x = 2, x = -1

7. № 400

   $$\frac{3}{x} - \frac{3}{x+4} = 1$$

   Решение:

   $$\frac{3(x+4) - 3x}{x(x+4)} = 1$$

   $$\frac{3x + 12 - 3x}{x^2 + 4x} = 1$$

   $$\frac{12}{x^2 + 4x} = 1$$

   $$12 = x^2 + 4x$$

   $$x^2 + 4x - 12 = 0$$

   $$D = 4^2 - 4 \cdot 1 \cdot (-12) = 16 + 48 = 64$$

   $$x_1 = \frac{-4 + \sqrt{64}}{2 \cdot 1} = \frac{-4 + 8}{2} = \frac{4}{2} = 2$$

   $$x_2 = \frac{-4 - \sqrt{64}}{2 \cdot 1} = \frac{-4 - 8}{2} = \frac{-12}{2} = -6$$

   Ответ: x = 2, x = -6