Задание 3. Найдите корень уравнения. 1) (x-5)² =(x-8)²; 2) (x+9)²=(x+6)²; 3) (x+10)²= (5-x)²; 4) (x-3)²=(x+10)²; 5) (x+6)²=(15-x)²; 6) (x-2)²=(x-9)².

Решение:

Применим формулу разности квадратов: a² - b² = (a - b)(a + b) = 0. Это значит, что либо (a - b) = 0, либо (a + b) = 0.

1. (x-5)² =(x-8)²
   (x-5)² - (x-8)² = 0
   ((x-5) - (x-8))((x-5) + (x-8)) = 0
   (x - 5 - x + 8)(x - 5 + x - 8) = 0
   (3)(2x - 13) = 0
   2x - 13 = 0
   2x = 13
   x = 6.5
2. (x+9)²=(x+6)²
   (x+9)² - (x+6)² = 0
   ((x+9) - (x+6))((x+9) + (x+6)) = 0
   (x + 9 - x - 6)(x + 9 + x + 6) = 0
   (3)(2x + 15) = 0
   2x + 15 = 0
   2x = -15
   x = -7.5
3. (x+10)²= (5-x)²
   (x+10)² - (5-x)² = 0
   ((x+10) - (5-x))((x+10) + (5-x)) = 0
   (x + 10 - 5 + x)(x + 10 + 5 - x) = 0
   (2x + 5)(15) = 0
   2x + 5 = 0
   2x = -5
   x = -2.5
4. (x-3)²=(x+10)²
   (x-3)² - (x+10)² = 0
   ((x-3) - (x+10))((x-3) + (x+10)) = 0
   (x - 3 - x - 10)(x - 3 + x + 10) = 0
   (-13)(2x + 7) = 0
   2x + 7 = 0
   2x = -7
   x = -3.5
5. (x+6)²=(15-x)²
   (x+6)² - (15-x)² = 0
   ((x+6) - (15-x))((x+6) + (15-x)) = 0
   (x + 6 - 15 + x)(x + 6 + 15 - x) = 0
   (2x - 9)(21) = 0
   2x - 9 = 0
   2x = 9
   x = 4.5
6. (x-2)²=(x-9)²
   (x-2)² - (x-9)² = 0
   ((x-2) - (x-9))((x-2) + (x-9)) = 0
   (x - 2 - x + 9)(x - 2 + x - 9) = 0
   (7)(2x - 11) = 0
   2x - 11 = 0
   2x = 11
   x = 5.5

Ответ: 6.5, -7.5, -2.5, -3.5, 4.5, 5.5