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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.
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Pertanyaan 1 dari 15
1. Pertanyaan
1 pointsHasil dari \(\dfrac{120 m^7 n^5}{6 m^2 n^{-2}}\) adalah …
Benar
\(\dfrac{\cancelto{20}{120} m^7 n^5}{\cancel{6} m^2 n^{-2}}\)
\(20 m^{7-2} n^{5-(-2)}\)
\(20 m^{5} n^{7}\)
Salah
\(\dfrac{\cancelto{20}{120} m^7 n^5}{\cancel{6} m^2 n^{-2}}\)
\(20 m^{7-2} n^{5-(-2)}\)
\(20 m^{5} n^{7}\)
Hint
\(\dfrac{a^m}{a^n} = a^{m-n}\) -
Pertanyaan 2 dari 15
2. Pertanyaan
1 pointsJika \(60^{10} = 2^x\:3^y\:5^z\) untuk \(x, y, z\) bilangan bulat, maka \(x + y -2z = \dotso\)
Benar
\((2^2\cdot 3\cdot 5)^{10} = 2^x\:3^y\:5^z\)
\(2^{20}\cdot 3^{10}\cdot 5^{10} = 2^x\:3^y\:5^z\)
\(x = 20\)
\(y = 10\)
\(z = 10\)
\(x + y -2z = 20 + 10 -2(10)\)
\(x + y -2z = 10\)
Salah
\((2^2\cdot 3\cdot 5)^{10} = 2^x\:3^y\:5^z\)
\(2^{20}\cdot 3^{10}\cdot 5^{10} = 2^x\:3^y\:5^z\)
\(x = 20\)
\(y = 10\)
\(z = 10\)
\(x + y -2z = 20 + 10 -2(10)\)
\(x + y -2z = 10\)
Hint
\((a^m)^n = a^{mn}\) -
Pertanyaan 3 dari 15
3. Pertanyaan
1 pointsBentuk \(5^{-\frac{3}{7}}\) sama dengan bentuk …
Benar
\(5^{-\frac{3}{7}} = \dfrac{1}{5^{\frac{3}{7}}}\)
\(5^{-\frac{3}{7}} = \dfrac{1}{\sqrt[7]{5^3}}\)
Salah
\(5^{-\frac{3}{7}} = \dfrac{1}{5^{\frac{3}{7}}}\)
\(5^{-\frac{3}{7}} = \dfrac{1}{\sqrt[7]{5^3}}\)
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Pertanyaan 4 dari 15
4. Pertanyaan
1 pointsJika \(\sqrt[3]{x^5\cdot \sqrt[6]{x^5}} = x^p\), maka nilai \(18p\) adalah …
Benar
\(\sqrt[3]{x^5\cdot x^{\frac{5}{6}}} = x^p\)
\(\sqrt[3]{x^{5+\frac{5}{6}}} = x^p\)
\(\sqrt[3]{x^{\frac{35}{6}}} = x^p\)
\(x^{\frac{35}{18}} = x^p\)
\(p=\frac{35}{18}\)
\(18p = \cancel{18}\times \frac{35}{\cancel{18}}\)
\(18p = 35\)
Salah
\(\sqrt[3]{x^5\cdot x^{\frac{5}{6}}} = x^p\)
\(\sqrt[3]{x^{5+\frac{5}{6}}} = x^p\)
\(\sqrt[3]{x^{\frac{35}{6}}} = x^p\)
\(x^{\frac{35}{18}} = x^p\)
\(p=\frac{35}{18}\)
\(18p = \cancel{18}\times \frac{35}{\cancel{18}}\)
\(18p = 35\)
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Pertanyaan 5 dari 15
5. Pertanyaan
1 pointsHasil dari \(\dfrac{5^{32}\times 5^{-7}}{5^{-1}}\) adalah …
Benar
\(\dfrac{5^{32-7}}{5^{-1}}\)
\(\dfrac{5^{25}}{5^{-1}}\)
\(5^{25-(-1)}\)
\(5^{26}\)
Salah
\(\dfrac{5^{32-7}}{5^{-1}}\)
\(\dfrac{5^{25}}{5^{-1}}\)
\(5^{25-(-1)}\)
\(5^{26}\)
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Pertanyaan 6 dari 15
6. Pertanyaan
1 pointsJika \(2^x + 2^{-x} = 8\), maka nilai dari \(4^x + 4^{-x} = \dotso\)
Benar
Kuadratkan kedua ruas:
\((2^x + 2^{-x})^2 = 8^2\)\(2^{2x} + 2\cdot 2^x\cdot 2^{-x} + 2^{-2x} = 64\)
\(4^{x} + 2 + 4^{-x} = 64\)
\(4^{x} + 4^{-x} = 64-2\)
\(4^{x} + 4^{-x} = 62\)
Salah
Kuadratkan kedua ruas:
\((2^x + 2^{-x})^2 = 8^2\)\(2^{2x} + 2\cdot 2^x\cdot 2^{-x} + 2^{-2x} = 64\)
\(4^{x} + 2 + 4^{-x} = 64\)
\(4^{x} + 4^{-x} = 64-2\)
\(4^{x} + 4^{-x} = 62\)
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Pertanyaan 7 dari 15
7. Pertanyaan
1 pointsJika \(3^x = a\), maka \(3^{2x + 2} = \dotso\)
Benar
\(3^{2x + 2} = 3^{2x}\cdot 3^2\)
\(3^{2x + 2} = (3^{x})^2\cdot 9\)
\(3^{2x + 2} = a^2\cdot 9\)
\(3^{2x + 2} = 9a^2\)
Salah
\(3^{2x + 2} = 3^{2x}\cdot 3^2\)
\(3^{2x + 2} = (3^{x})^2\cdot 9\)
\(3^{2x + 2} = a^2\cdot 9\)
\(3^{2x + 2} = 9a^2\)
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Pertanyaan 8 dari 15
8. Pertanyaan
1 pointsHasil dari \(\dfrac{5^{n + 3} -5^{n + 1}}{5^{n + 1} + 5^{n + 2}}\) adalah …
Benar
\(\dfrac{5^n\cdot 5^3 -5^n\cdot 5^1}{5^n\cdot 5^1 + 5^n\cdot 5^2}\)
\(\dfrac{5^n\cdot 125 -5^n\cdot 5}{5^n\cdot 5 + 5^n\cdot 25}\)
\(\dfrac{\cancel{5^n}(125-5)}{\cancel{5^n}(5 + 25)}\)
\(\dfrac{120}{30}\)
\(4\)
Salah
\(\dfrac{5^n\cdot 5^3 -5^n\cdot 5^1}{5^n\cdot 5^1 + 5^n\cdot 5^2}\)
\(\dfrac{5^n\cdot 125 -5^n\cdot 5}{5^n\cdot 5 + 5^n\cdot 25}\)
\(\dfrac{\cancel{5^n}(125-5)}{\cancel{5^n}(5 + 25)}\)
\(\dfrac{120}{30}\)
\(4\)
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Pertanyaan 9 dari 15
9. Pertanyaan
1 pointsHasil dari \(5^0 + 5^2 -5^{-1}\) adalah …
Benar
\(5^0 + 5^2 -5^{-1} = 1 + 25 -\frac{1}{5}\)
\(5^0 + 5^2 -5^{-1} = 26 -0,2\)
\(5^0 + 5^2 -5^{-1} = 25,80\)
Salah
\(5^0 + 5^2 -5^{-1} = 1 + 25 -\frac{1}{5}\)
\(5^0 + 5^2 -5^{-1} = 26 -0,2\)
\(5^0 + 5^2 -5^{-1} = 25,80\)
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Pertanyaan 10 dari 15
10. Pertanyaan
1 pointsBentuk sederhana dari \(\dfrac{x^{-1}-y^{-1}}{x-y}\) adalah …
Benar
\(\dfrac{x^{-1}-y^{-1}}{x-y}\)
\(\dfrac{\frac{1}{x}-\frac{1}{y}}{x-y}\)
\(\dfrac{\frac{\cancel{y-x}}{xy}}{-\cancel{(y-x)}}\)
\(-\dfrac{1}{xy}\)
Salah
\(\dfrac{x^{-1}-y^{-1}}{x-y}\)
\(\dfrac{\frac{1}{x}-\frac{1}{y}}{x-y}\)
\(\dfrac{\frac{\cancel{y-x}}{xy}}{-\cancel{(y-x)}}\)
\(-\dfrac{1}{xy}\)
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Pertanyaan 11 dari 15
11. Pertanyaan
1 pointsHasil dari \(2\sqrt{48} + \sqrt{12}-\sqrt{75}-\sqrt{3}\) adalah …
Benar
\(2\sqrt{16\cdot 3} + \sqrt{4\cdot 3}-\sqrt{25\cdot 3}-\sqrt{3}\)
\(2\cdot 4\sqrt{3} + 2\sqrt{3}-5\sqrt{3}-\sqrt{3}\)
\(8\sqrt{3} + 2\sqrt{3}-5\sqrt{3}-\sqrt{3}\)
\((8 + 2-5-1)\sqrt{3}\)
\(4\sqrt{3}\)
Salah
\(2\sqrt{16\cdot 3} + \sqrt{4\cdot 3}-\sqrt{25\cdot 3}-\sqrt{3}\)
\(2\cdot 4\sqrt{3} + 2\sqrt{3}-5\sqrt{3}-\sqrt{3}\)
\(8\sqrt{3} + 2\sqrt{3}-5\sqrt{3}-\sqrt{3}\)
\((8 + 2-5-1)\sqrt{3}\)
\(4\sqrt{3}\)
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Pertanyaan 12 dari 15
12. Pertanyaan
1 pointsBentuk sederhana dari \(\dfrac{3\sqrt{2}-3}{\sqrt{2} + 1}\) adalah …
Benar
\(\dfrac{3\sqrt{2}-3}{\sqrt{2} + 1}\times \color{red} \dfrac{\sqrt{2}-1}{\sqrt{2}-1}\)
\(\dfrac{(3\sqrt{2}-3)(\sqrt{2}-1)}{2-1}\)
\(\dfrac{6-3\sqrt{2}-3\sqrt{2} + 3}{1}\)
\(9-6\sqrt{2}\)
Salah
\(\dfrac{3\sqrt{2}-3}{\sqrt{2} + 1}\times \color{red} \dfrac{\sqrt{2}-1}{\sqrt{2}-1}\)
\(\dfrac{(3\sqrt{2}-3)(\sqrt{2}-1)}{2-1}\)
\(\dfrac{6-3\sqrt{2}-3\sqrt{2} + 3}{1}\)
\(9-6\sqrt{2}\)
Hint
Kalikan dengan bentuk sekawan \(\color{red} \frac{\sqrt{2}\:-\:1}{\sqrt{2}\:-\:1}\) -
Pertanyaan 13 dari 15
13. Pertanyaan
1 pointsHasil dari \(\sqrt{12 + 2\sqrt{32}}-\sqrt{9-2\sqrt{8}} = \dotso\)
Benar
\(\sqrt{12 + 2\sqrt{32}} = \sqrt{(8 + 4) + 2\sqrt{8\cdot4}}\)
\(\sqrt{12 + 2\sqrt{32}} = \sqrt{8} + \sqrt{4}\)
\(\sqrt{12 + 2\sqrt{32}} = 2\sqrt{2} + 2\)
\(\sqrt{9-2\sqrt{8}} = \sqrt{(8 + 1) -2\sqrt{8\cdot 1}}\)
\(\sqrt{9-2\sqrt{8}} = \sqrt{8}-1\)
\(\sqrt{9-2\sqrt{8}} = 2\sqrt{2}-1\)
Jadi,
\(\sqrt{12 + 2\sqrt{32}}-\sqrt{9-2\sqrt{8}}\)
\(2\sqrt{2} + 2-(2\sqrt{2}-1)\)
\(3\)
Salah
\(\sqrt{12 + 2\sqrt{32}} = \sqrt{(8 + 4) + 2\sqrt{8\cdot4}}\)
\(\sqrt{12 + 2\sqrt{32}} = \sqrt{8} + \sqrt{4}\)
\(\sqrt{12 + 2\sqrt{32}} = 2\sqrt{2} + 2\)
\(\sqrt{9-2\sqrt{8}} = \sqrt{(8 + 1) -2\sqrt{8\cdot 1}}\)
\(\sqrt{9-2\sqrt{8}} = \sqrt{8}-1\)
\(\sqrt{9-2\sqrt{8}} = 2\sqrt{2}-1\)
Jadi,
\(\sqrt{12 + 2\sqrt{32}}-\sqrt{9-2\sqrt{8}}\)
\(2\sqrt{2} + 2-(2\sqrt{2}-1)\)
\(3\)
Hint
Gunakan rumus penarikan akar: \(\sqrt{(a + b)\pm 2\sqrt{a\cdot b}} = \sqrt{a} \pm \sqrt{b}\) \(a > b\) -
Pertanyaan 14 dari 15
14. Pertanyaan
1 pointsBentuk sederhana dari \(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3}-\sqrt{2}} + \sqrt{54}-\sqrt{150}\) adalah …
Benar
\(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3}-\sqrt{2}}\color{red}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \color{black}+ \sqrt{9\cdot 6}-\sqrt{25\cdot 6}\)
\(\dfrac{3 + 2\sqrt{6} + 2}{3-2} + 3\sqrt{6}-5\sqrt{6}\)
\(\dfrac{5 + 2\sqrt{6}}{1} + 3\sqrt{6}-5\sqrt{6}\)
\(5 + 2\sqrt{6} + 3\sqrt{6}-5\sqrt{6}\)
\(5\)
Salah
\(\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3}-\sqrt{2}}\color{red}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \color{black}+ \sqrt{9\cdot 6}-\sqrt{25\cdot 6}\)
\(\dfrac{3 + 2\sqrt{6} + 2}{3-2} + 3\sqrt{6}-5\sqrt{6}\)
\(\dfrac{5 + 2\sqrt{6}}{1} + 3\sqrt{6}-5\sqrt{6}\)
\(5 + 2\sqrt{6} + 3\sqrt{6}-5\sqrt{6}\)
\(5\)
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Pertanyaan 15 dari 15
15. Pertanyaan
1 pointsJika \(p = \frac{1}{27}\) dan \(q = 16\) maka nilai dari \(\dfrac{p^{-\frac{1}{3}}-q^{\frac{3}{4}}}{p^{-\frac{2}{3}}+ q^{-\frac{1}{2}}}=\dotso\)
Benar
\(p = 3^{-3}\)
\(q = 2^4\)
\(\dfrac{(3^{-3})^{-\frac{1}{3}}-(2^4)^{\frac{3}{4}}}{(3^{-3})^{-\frac{2}{3}}+ (2^4)^{-\frac{1}{2}}}\)
\(\dfrac{3-2^3}{3^2 + 2^{-2}}\)
\(\dfrac{3-8}{9 + \frac{1}{4}}\)
\(\dfrac{-5}{\frac{37}{4}}\)
\(-\frac{20}{37}\)
Salah
\(p = 3^{-3}\)
\(q = 2^4\)
\(\dfrac{(3^{-3})^{-\frac{1}{3}}-(2^4)^{\frac{3}{4}}}{(3^{-3})^{-\frac{2}{3}}+ (2^4)^{-\frac{1}{2}}}\)
\(\dfrac{3-2^3}{3^2 + 2^{-2}}\)
\(\dfrac{3-8}{9 + \frac{1}{4}}\)
\(\dfrac{-5}{\frac{37}{4}}\)
\(-\frac{20}{37}\)