Quiz-summary
0 of 4 questions completed
Pertanyaan:
- 1
- 2
- 3
- 4
Information
Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 60 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
Anda telah menyelesaikan kuis sebelumnya. Oleh karena itu, Anda tidak dapat memulainya lagi.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Hasil
0 dari 4 pertanyaan terjawab dengan benar
Waktu Anda:
Time has elapsed
Anda telah meraih 0 dari 0 poin, (0)
Categories
- Not categorized 0%
- 1
- 2
- 3
- 4
- Dijawab
- Ragu-Ragu
-
Pertanyaan 1 dari 4
1. Pertanyaan
1 pointsHimpunan penyelesaian persamaan \((2-\sqrt{3})^{x^2+x-6}+(2+\sqrt{3})^{x^2+x-6}-2 = 0\) adalah …
Benar
\(\color{blue} 2 -\sqrt{3} = \frac{1}{2+\sqrt{3}}\)
\((2-\sqrt{3})^{x^2 + x – 6} + (2+\sqrt{3})^{x^2 + x -6}- 2 = 0\)
\((\frac{1}{2+\sqrt{3}})^{x^2+ x-6}+ (2+\sqrt{3})^{x^2+x-6}- 2 = 0\)
Misal \(p = (2+\sqrt{3})^{x^2 +x- 6}\)
\(\frac{1}{p}+p-2 = 0\:\:\:\:\:\color{blue}\text{kali }p\)
\(1+p^2-2p = 0\)
\(p^2-2p+1= 0\)
\((p-1)(p-1) = 0\)
\(p = 1\)
\((2+\sqrt{3})^{x^2+x- 6} = 1\)
Solusi:
\(x^2+x-6 = 0\)
\((x+3)(x-2) = 0\)
\(x+3 = 0 \rightarrow x_1 = -3\)
\(x -2 = 0 \rightarrow x_2 = 2\)
Himpunan penyelesaian adalah {-3, 2}
Salah
\(\color{blue} 2 -\sqrt{3} = \frac{1}{2+\sqrt{3}}\)
\((2-\sqrt{3})^{x^2 + x – 6} + (2+\sqrt{3})^{x^2 + x -6}- 2 = 0\)
\((\frac{1}{2+\sqrt{3}})^{x^2+ x-6}+ (2+\sqrt{3})^{x^2+x-6}- 2 = 0\)
Misal \(p = (2+\sqrt{3})^{x^2 +x- 6}\)
\(\frac{1}{p}+p-2 = 0\:\:\:\:\:\color{blue}\text{kali }p\)
\(1+p^2-2p = 0\)
\(p^2-2p+1= 0\)
\((p-1)(p-1) = 0\)
\(p = 1\)
\((2+\sqrt{3})^{x^2+x- 6} = 1\)
Solusi:
\(x^2+x-6 = 0\)
\((x+3)(x-2) = 0\)
\(x+3 = 0 \rightarrow x_1 = -3\)
\(x -2 = 0 \rightarrow x_2 = 2\)
Himpunan penyelesaian adalah {-3, 2}
-
Pertanyaan 2 dari 4
2. Pertanyaan
1 pointsJumlah semua nilai \(x\) yang memenuhi persamaan \(5^{2x^3 – 5x^2 – 3x} = 1\) adalah …
Benar
\(5^{2x^3 – 5x^2 – 3x} = 1\)
Solusi:
\(2x^3 – 5x^2 – 3x = 0\)
\(x(2x^2 – 5x – 3) = 0\)
\(x(2x + 1)(x – 3) = 0\)
\(x_1 = 0\)
\(2x + 1 = 0 \rightarrow x_2 = -\frac{1}{2}\)
\(x – 3 = 0 \rightarrow x_3 = 3\)
Jumlah semua nilai \(x\) adalah \(0 -\frac{1}{2} + 3 = 2,5\)
Salah
\(5^{2x^3 – 5x^2 – 3x} = 1\)
Solusi:
\(2x^3 – 5x^2 – 3x = 0\)
\(x(2x^2 – 5x – 3) = 0\)
\(x(2x + 1)(x – 3) = 0\)
\(x_1 = 0\)
\(2x + 1 = 0 \rightarrow x_2 = -\frac{1}{2}\)
\(x – 3 = 0 \rightarrow x_3 = 3\)
Jumlah semua nilai \(x\) adalah \(0 -\frac{1}{2} + 3 = 2,5\)
-
Pertanyaan 3 dari 4
3. Pertanyaan
1 pointsHimpunan penyelesaian \(7^{4x^2 – 11x – 1} = 49\) adalah …
Benar
\(7^{4x^2 – 11x – 1} = 7^2\)
\(4x^2 – 11x – 1 = 2\)
\(4x^2 – 11x – 3 = 0\)
\((4x + 1)(x – 3) = 0\)
\(4x + 1 = 0 \rightarrow x_1 = -\frac{1}{4}\)
\(x – 3 = 0 \rightarrow x_2 = 3\)
Himpunan penyelesaiannya adalah \(\lbrace -\frac{1}{4}, 3 \rbrace\)
Salah
\(7^{4x^2 – 11x – 1} = 7^2\)
\(4x^2 – 11x – 1 = 2\)
\(4x^2 – 11x – 3 = 0\)
\((4x + 1)(x – 3) = 0\)
\(4x + 1 = 0 \rightarrow x_1 = -\frac{1}{4}\)
\(x – 3 = 0 \rightarrow x_2 = 3\)
Himpunan penyelesaiannya adalah \(\lbrace -\frac{1}{4}, 3 \rbrace\)
-
Pertanyaan 4 dari 4
4. Pertanyaan
1 pointsHimpunan bilangan bulat yang merupakan solusi persamaan \((x + 2)^{x^2 – 3x – 4} = (2x – 3)^{x^2 – 3x – 4}\) adalah …
Benar
\((x + 2)^{x^2 – 3x – 4} = (2x – 3)^{x^2 – 3x – 4}\)
Bentuk \(f(x)^{h(x)} = g(x)^{h(x)}\)
Solusi 1: \(f(x) = g(x)\)
\(x + 2 = 2x – 3\)
\(2 + 3 = 2x – x\)
\(x_1 = 5\)
Solusi 2: \(h(x) = 0\) syarat \(f(x) \neq 0, g(x) \neq 0\)
\(x^2 – 3x – 4 = 0\)
\((x – 4)(x + 1) = 0\)
\(x – 4 = 0 \rightarrow x_2 = 4\)
\(x + 1 = 0 \rightarrow x_3 = -1\)
Solusi 3: \(f(x) = -g(x)\) syarat \(h(x)\) genap
\(x + 2 = – (2x – 3)\)
\(x + 2 = – 2x + 3\)
\(3x = 1\)
\(x = \frac{1}{3}\)
Tidak diambil karena bukan bilangan bulat
Jadi solusinya adalah {−1, 4, 5}
Salah
\((x + 2)^{x^2 – 3x – 4} = (2x – 3)^{x^2 – 3x – 4}\)
Bentuk \(f(x)^{h(x)} = g(x)^{h(x)}\)
Solusi 1: \(f(x) = g(x)\)
\(x + 2 = 2x – 3\)
\(2 + 3 = 2x – x\)
\(x_1 = 5\)
Solusi 2: \(h(x) = 0\) syarat \(f(x) \neq 0, g(x) \neq 0\)
\(x^2 – 3x – 4 = 0\)
\((x – 4)(x + 1) = 0\)
\(x – 4 = 0 \rightarrow x_2 = 4\)
\(x + 1 = 0 \rightarrow x_3 = -1\)
Solusi 3: \(f(x) = -g(x)\) syarat \(h(x)\) genap
\(x + 2 = – (2x – 3)\)
\(x + 2 = – 2x + 3\)
\(3x = 1\)
\(x = \frac{1}{3}\)
Tidak diambil karena bukan bilangan bulat
Jadi solusinya adalah {−1, 4, 5}