\((x\:-\:2)^{x^2\:-\: x \:-\:2} = 1\)
Langkah 1: \(f(x) = 1\)
\(x\:-\:2 = 1\)
\(x = 1 +2\)
\(x = 3\)
Langkah 2: \(f(x) = -1\), dengan \(g(x)\) harus genap
\(x\:-\:2 = -1\)
\(x = – 1 +2\)
\(x = 1\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(g(1) = 1^2 \:-\: 1 \:-\:2 = -2\:\:\:\:\:\color{blue}\text{genap}\)
Langkah 3: \(g(x) = 0, f(x) \neq 0\)
\(x^2\:-\: x \:-\: 2 = 0\)
\((x\:-\:2)(x + 1) = 0\)
\(x = 2\:\:\:\:\:\color{red}\text{tidak memenuhi}\)
\(\color{red} f(2) = 2- 2 = 0\)
\(x = -1\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(f(-1) = -1\:-\: 2 \neq 0\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -1, \: 1, \: 3\rbrace}\)
Langkah 1: \(f(x) = 1\)
\(x^2 + 4x + 5 = 1\)
\(x^2 + 4x + 4 = 0\)
\((x + 2)^2 = 0\)
\(x + 2 = 0\)
\(x = -2\)
Langkah 2: \(f(x) = -1\), dengan \(g(x)\) harus genap
\(x^2 + 4x + 5 = -1\)
\(x^2 + 4x + 6 = 0\)
\(\text{Cek nilai diskriminan:}\)
\(b^2 \:-\: 4ac = 4^2\:-\: 4(1)(6) < 0\:\:\:\:\:\color{red} \text{tidak ada solusi}\)
Langkah 3: \(g(x) = 0, f(x) \neq 0\)
\(x^2 \:-\: 25 = 0\)
\((x + 5)(x \:-\: 5) = 0\)
\(x_1 = -5\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(f(-5) = (-5)^2 + 4(-5) + 5 \neq 0\)
\(x_2 = 5\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(f(5) = (5)^2 + 4(5) + 5 \neq 0\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -5, \:-2, \:5\rbrace}\)
Tentukan himpunan penyelesaian persamaan \((2x + 7)^{2x\:-\:4}\:-\:2(2x + 7)^{x\:-\:2} + 1 = 0\)
\((2x + 7)^{2(x\:-\:2)}\:-\:2(2x + 7)^{x\:-\:2} + 1 = 0\)
Misal \(p = (2x + 7)^{x\:-\:2}\)
\(p^2\: -\: 2p + 1 = 0\)
\((p\: -\: 1)^2 =0\)
\(p = 1\)
\((2x + 7)^{x\:-\:2} = 1\)
Langkah 1: \(f(x) = 1\)
\(2x + 7 = 1\)
\(2x = 1\: -\: 7\)
\(2x = – 6\)
\(x = -\dfrac{6}{2} = – 3\)
Langkah 2: \(f(x) = -1\), dengan \(g(x)\) harus genap
\(2x + 7 = -1\)
\(2x = – 1 \:-\: 7\)
\(2x = -8\)
\(x = -\dfrac{8}{2} = -4\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(g(-2) = (-4) – 2 = -6\:\:\:\:\:\color{blue}\text{genap}\)
Langkah 3: \(g(x) = 0, f(x) \neq 0\)
\(x \:-\:2 = 0\)
\(x = 2\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(f(2) = 2(2) + 7 \neq 0\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -4, \: -3, \: 2\rbrace}\)
Langkah 1: \(f(x) = 1\)
\(2x + 11 = 1\)
\(2x = 1 \:-\:11\)
\(2x = – 10\)
\(x = -5\)
Langkah 2: \(f(x) = -1\), dengan \(g(x)\) harus genap
\(2x + 11 = -1\)
\(2x = – 1 \:-\:11\)
\(2x = -12\)
\(x = -\dfrac{12}{2} = -6\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(g(-6) = (-6) – 2 = -8\:\:\:\:\:\color{blue}\text{genap}\)
Langkah 3: \(g(x) = 0, f(x) \neq 0\)
\(x \:-\: 2 = 0\)
\(x = 2\:\:\:\:\:\color{blue}\text{memenuhi}\)
\(f(2) = 2(2) + 11 \neq 0\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -6, \: -5, \: 2\rbrace}\)
