Решите уравнение:
а) $(x - 1)(x + 1) = 2(x - 3)^2 - x^2$;
б) $(2x + 3)^2 - 4(x - 1)(x + 1) = 49$;
в) $3(x + 5)^2 - 4x^2 = (2 - x)(2 + x)$;
г) $(3x + 1)^2 - (3x - 2)(2 + 3x) = 17$.
$(x - 1)(x + 1) = 2(x - 3)^2 - x^2$
$x^2 - 1 = 2(x^2 - 6x + 9) - x^2$
$x^2 - 1 = 2x^2 - 12x + 18 - x^2$
$x^2 + 12x - 2x^2 + x^2 = 18 + 1$
12x = 19
$x = \frac{19}{12} = 1\frac{7}{12}$
$(2x + 3)^2 - 4(x - 1)(x + 1) = 49$
$4x^2 + 12x + 9 - 4(x^2 - 1) = 49$
$4x^2 + 12x + 9 - 4x^2 + 4 = 49$
12x = 49 − 9 − 4
12x = 36
x = 3
$3(x + 5)^2 - 4x^2 = (2 - x)(2 + x)$
$3(x^2 + 10x + 25) - 4x^2 = 4 - x^2$
$3x^2 + 30x + 75 - 4x^2 + x^2 = 4$
30x = 4 − 75
30x = −71
$x = -\frac{71}{30} = -2\frac{11}{30}$
$(3x + 1)^2 - (3x - 2)(2 + 3x) = 17$
$(3x + 1)^2 - (3x - 2)(3x + 2) = 17$
$9x^2 + 6x + 1 - (9x^2 - 4) = 17$
$9x^2 + 6x + 1 - 9x^2 + 4 = 17$
6x = 17 − 5
6x = 12
x = 2
Пожауйста, оцените решение
