|x^2+x+1|=|2x^2+x-1|

解答

x^{2} + x + 1 = 2 x^{2} + x - 1

x = - 2 or x = - \frac{2}{3} or x = 0 or x = 2

十进制

x = - 1.41421 \dots, x = - 0.66666 \dots, x = 0, x = 1.41421 \dots

求解步骤

x^{2} + x + 1 = 2 x^{2} + x - 1

找到正负区间

x \leq - 1, - 1 < x < \frac{1}{2}, x \geq \frac{1}{2}

对每个区间解不等式

x = - 2 or (x = - \frac{2}{3} or x = 0) or x = 2

x = - 2 or x = - \frac{2}{3} or x = 0 or x = 2

∣ 11 x + 22 ∣ + ∣ 3 x - 1 ∣ = 6 x + 6

x^{2} - 3 \cdot x ∣ 2 ∣ = x^{2} - 2 x

2 ∣ x + 1 ∣ + ∣ x - 2 ∣ = 5

so l v e f or y, ∣ y - 1 ∣ = x

∣ 2 x - 3 ∣ + ∣ x + 2 ∣ = 4