[tex](x - a)(x - b) = (x - c)(x - d) \\ {x}^{2} - (a + b)x + ab = {x}^{2} - (c + d)x + cd \\ (a + b - c - d)x = ab - cd \\ x = \frac{ab - cd}{a + b - c - d} [/tex]
Преобразования равенства:
[tex]a + d = b + c \\ a - b = c - d \\ {a}^{2} + {b}^{2} - 2ab = {c}^{2} + {d}^{2} - 2cd \\ {a}^{2} + {b}^{2} - {c}^{2} - {d}^{2} = 2(ab - cd)[/tex]
Подстановка в первое равенство:
[tex]x = \frac{ {a}^{2} + {b}^{2} - {c}^{2} - {d}^{2} }{2(a + b - c - d)} \\ x = \frac{(a - d)(a + d) + (b - c)(b + c)}{2(a + b - c - d)} \\ x = \frac{2022(a - c + b - d)}{2(a - c + b - d)} = 1011[/tex]
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Verified answer
Ответ:
1011
Объяснение:
[tex](x - a)(x - b) = (x - c)(x - d) \\ {x}^{2} - (a + b)x + ab = {x}^{2} - (c + d)x + cd \\ (a + b - c - d)x = ab - cd \\ x = \frac{ab - cd}{a + b - c - d} [/tex]
Преобразования равенства:
[tex]a + d = b + c \\ a - b = c - d \\ {a}^{2} + {b}^{2} - 2ab = {c}^{2} + {d}^{2} - 2cd \\ {a}^{2} + {b}^{2} - {c}^{2} - {d}^{2} = 2(ab - cd)[/tex]
Подстановка в первое равенство:
[tex]x = \frac{ {a}^{2} + {b}^{2} - {c}^{2} - {d}^{2} }{2(a + b - c - d)} \\ x = \frac{(a - d)(a + d) + (b - c)(b + c)}{2(a + b - c - d)} \\ x = \frac{2022(a - c + b - d)}{2(a - c + b - d)} = 1011[/tex]
Корень всегда будет равняться 1011