一元二次方程的题.一共四个.用公式法解:1/2(x+1)²-4=0以下这几个用什么方法解都行,只要是一元二次方程的解法.

1个回答

  • ① (1 / 2)(x + 1)² - 4 = 0

    x ² + 2 x + 1 - 4 = 0

    x ² + 2 x - 3 = 0

    a = 1 ,b = 2 ,c = - 3

    ∵ △ = b ² - 4 a c

    = 2 ² - 4 × 1 × (- 3)

    = 4 + 4 × 3

    = 4 + 12

    = 16 > 0

    ∴ 此方程有两个不等实数根.

    ∴ x = 【 - b ± √(b ² - 4 a c)】 / 2 a

    = 【 - 2 ± √ 16 】 / 2 × 1

    = (- 2 ± 4)/ 2

    = - 1 ± 2

    x1 = - 1 + 2 = 1

    x2 = - 1 - 2 = - 3

    ② x ² - 2 x - 4 = 0

    x ² - 2 x + 1 = 4 + 1

    (x + 1)² = 5

    x + 1 = ±√5

    x = ±√5 - 1

    x1 = √5 - 1

    x2 = - √5 - 1

    ③ (x - 1)(x + 4)= 6

    x ² + 3 x - 4 = 6

    x ² + 3 x - 10 = 0

    (x + 5)(x - 2)= 0

    x + 5 = 0 或 x - 2 = 0

    x1 = - 5 x2 = 2

    ④ (x + 1)² - 4 = 0

    (x + 1 + 4)(x + 1 - 4)= 0

    (x + 5)(x - 3)= 0

    x + 5 = 0 或 x - 3 = 0

    x1 = - 5 x2 = 3

    ⑤ (x - 3)² - 2 x(x - 3)= 0

    (x - 3)(x - 3 - 2 x)= 0

    (x - 3)(- x - 3)= 0

    -(x - 3)(x + 3)= 0

    x - 3 = 0 或 x + 3 = 0

    x1 = 3 x2 = - 3

    ⑥ (1 / 2)(x + 1)² - 4 = 0

    (x + 1)² = 4

    x + 1 = ± 2

    x = ± 2 - 1

    x1 = 2 - 1 = 1

    x2 = - 2 - 1 = - 3

    ⑦ 方程与 ③ 重复.