|x+1||x+4|=1

解

∣ x + 1 ∣ ∣ x + 4 ∣ = 1

x = \frac{- 5 - 13}{2} or x = - \frac{5 + 5}{2} or x = - \frac{5 - 5}{2} or x = \frac{- 5 + 13}{2}

十進法表記

x = - 4.30277 \dots, x = - 3.61803 \dots, x = - 1.38196 \dots, x = - 0.69722 \dots

解答ステップ

∣ x + 1 ∣ ∣ x + 4 ∣ = 1

正と負の区間を求める

x < - 4, - 4 \leq x < - 1, x \geq - 1

各区間の不等式を解く

x = \frac{- 5 - 13}{2} or (x = - \frac{5 + 5}{2} or x = - \frac{5 - 5}{2}) or x = \frac{- 5 + 13}{2}

x = \frac{- 5 - 13}{2} or x = - \frac{5 + 5}{2} or x = - \frac{5 - 5}{2} or x = \frac{- 5 + 13}{2}