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Answer

−2<x<17ataux>1-2 < x < \dfrac{1}{7} \quad \text{atau} \quad x > 1

Step-by-step solution

8 steps
  1. Tentukan domain ketidaksamaan.

    Pembilang tidak dibatasi, tetapi penyebut tidak boleh nol:

    x+2≠0⇒x≠−2x + 2 \neq 0 \quad \Rightarrow \quad x \neq -2 x−1≠0⇒x≠1x - 1 \neq 0 \quad \Rightarrow \quad x \neq 1

    Domain: semua xx kecuali x=−2x = -2 dan x=1x = 1.

  2. Pindahkan ke satu sisi.

    x−3x+2−x+1x−1<0\frac{x-3}{x+2} - \frac{x+1}{x-1} < 0
  3. Penyebut bersama (x+2)(x−1)(x+2)(x-1).

    (x−3)(x−1)−(x+1)(x+2)(x+2)(x−1)<0\frac{(x-3)(x-1) - (x+1)(x+2)}{(x+2)(x-1)} < 0
  4. Hitung pembilang.

    (x−3)(x−1)=x2−4x+3(x-3)(x-1) = x^2 - 4x + 3
    (x+1)(x+2)=x2+3x+2(x+1)(x+2) = x^2 + 3x + 2
    Pembilang: x2−4x+3−(x2+3x+2)=x2−4x+3−x2−3x−2=−7x+1x^2 - 4x + 3 - (x^2 + 3x + 2) = x^2 - 4x + 3 - x^2 - 3x - 2 = -7x + 1

    1−7x(x+2)(x−1)<0\frac{1 - 7x}{(x+2)(x-1)} < 0
  5. Titik kritis.

    Pembilang =0= 0: 1−7x=0⇒x=171 - 7x = 0 \quad \Rightarrow \quad x = \frac{1}{7}
    Penyebut =0= 0: x=−2x = -2, x=1x = 1
    Titik kritis: x=−2x = -2, x=17x = \frac{1}{7}, x=1x = 1.

  6. Interval uji

    (−∞,−2)(-\infty, -2), (−2,17)(-2, \frac{1}{7}), (17,1)(\frac{1}{7}, 1), (1,∞)(1, \infty).

  7. Uji tanda.

    • Interval (−∞,−2)(-\infty, -2), uji x=−3x = -3:
      Pembilang: 1−7(−3)=22>01 - 7(-3) = 22 > 0
      Penyebut: (−3+2)(−3−1)=(−1)(−4)=4>0(-3+2)(-3-1) = (-1)(-4) = 4 > 0
      Fraksi: ++>0\frac{+}{+} > 0 (tidak memenuhi).

    • Interval (−2,17)(-2, \frac{1}{7}), uji x=0x = 0:
      Pembilang: 1−7(0)=1>01 - 7(0) = 1 > 0
      Penyebut: (0+2)(0−1)=(2)(−1)=−2<0(0+2)(0-1) = (2)(-1) = -2 < 0
      Fraksi: +−<0\frac{+}{-} < 0 (memenuhi).

    • Interval (17,1)(\frac{1}{7}, 1), uji x=0.5x = 0.5:
      Pembilang: 1−7(0.5)=−2.5<01 - 7(0.5) = -2.5 < 0
      Penyebut: (0.5+2)(0.5−1)=(2.5)(−0.5)=−1.25<0(0.5+2)(0.5-1) = (2.5)(-0.5) = -1.25 < 0
      Fraksi: −−>0\frac{-}{-} > 0 (tidak memenuhi).

    • Interval (1,∞)(1, \infty), uji x=2x = 2:
      Pembilang: 1−7(2)=−13<01 - 7(2) = -13 < 0
      Penyebut: (2+2)(2−1)=(4)(1)=4>0(2+2)(2-1) = (4)(1) = 4 > 0
      Fraksi: −+<0\frac{-}{+} < 0 (memenuhi).

  8. Solusi ketat <0< 0, tidak termasuk x=17x = \frac{1}{7}.

    Jawaban:
    -2 < x < \frac{1}{7} atau x > 1

    −2<x<17ataux>1-2 < x < \dfrac{1}{7} \quad \text{atau} \quad x > 1
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