Operations on real numbers and the laws of exponents

Correct answer: A. (30)

Explanation: Here (5^{-2}+5^{-3}=\frac{1}{25}+\frac{1}{125}=\frac{6}{125}), and (5^{-4}=\frac{1}{625}). Division gives (30).

Correct answer: C. 17

Explanation: Using the identity \((a-b)^2=a^2+b^2-2ab\), we get \(x^2=(\sqrt{11}-\sqrt{6})^2=11+6-2\sqrt{66}=17-2\sqrt{66}\). Therefore, \(x^2+2\sqrt{66}=17\). Option 5 results from incorrectly subtracting 6 from 11 while squaring. Exam tip: for \((a-b)^2\), the middle term is always \(-2ab\).

Correct answer: A. \(9r^{6}s^{-8}\)

Explanation: Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).

Correct answer: A. (\frac{9}{2})

Explanation: Since (1024=2^{10}), (16^{x}=2^{4x}) gives (x=\frac{5}{2}), and (32^{y}=2^{5y}) gives (y=2). Hence the sum is (\frac{9}{2}).

Correct answer: A. \(a^p\cdot a^q=a^{p+q}\)

Explanation: When powers with the same positive base are multiplied, their exponents are added; hence \(a^p\cdot a^q=a^{p+q}\). In division, exponents are subtracted, not divided: \(a^p/a^q=a^{p-q}\). Exam tip: multiply → add exponents; divide → subtract.

Correct answer: B. (6)

Explanation: Since (24^{3}=(2^{3}\cdot3)^{3}=2^{9}\cdot3^{3}), division leaves (2^{3}\cdot3=24), so the correct value is not among the options.

Correct answer: A. (16\sqrt{17})

Explanation: Here (\frac{1}{s}=\sqrt{17}-4), so (s-\frac{1}{s}=8) and (s+\frac{1}{s}=2\sqrt{17}). Thus (s^{2}-\frac{1}{s^{2}}=16\sqrt{17}).

Correct answer: A. \(\frac{4x^{3}y^{4}}{5}\)

Explanation: We get \(\left(\frac{125x^{-9}}{64y^{12}}\right)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}\). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).

Correct answer: C. (10)

Explanation: We use \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\). Thus \(60=6\left(x+\frac{1}{x}\right)\), so the value is (10).

Correct answer: A. \(5b^{2}\)

Explanation: Combining like terms gives \(6b^{-3}+9b^{-3}=15b^{-3}\). Therefore, \(\frac{15b^{-3}}{3b^{-5}}=5b^{-3-(-5)}=5b^2\), so option A is correct. Option B results from subtracting the exponents in the wrong order. Exam tip: when dividing powers with the same non-zero base, subtract the denominator exponent from the numerator exponent: \(b^m/b^n=b^{m-n}\).

Correct answer: A. (250\sqrt{2})

Explanation: From (\sqrt{x}=5\sqrt{2}), (x=50), and (x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).

Correct answer: A. \(\frac{4721}{1600}\)

Explanation: Here \(\left(\frac{5}{8}\right)^{-2}=\frac{64}{25}\) and \(\left(\frac{8}{5}\right)^{-2}=\frac{25}{64}\). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).

Correct answer: A. 2

Explanation: Using the laws of exponents, \\( (7^x)^2=7^{2x}\\), so the left-hand side becomes \\(7^{2x}\cdot7^{x-1}=7^{3x-1}\\). Also, \\(16807=7^5\\). Therefore, \\(3x-1=5\\), giving \\(3x=6\\) and hence \\(x=2\\). Exam tip: when multiplying powers with the same base, add their exponents.

Correct answer: C. (15)

Explanation: Here (\sqrt{363}=11\sqrt{3}), (2\sqrt{147}=14\sqrt{3}), and (3\sqrt{75}=15\sqrt{3}). The numerator is (12\sqrt{3}), so the value should be (12).

Correct answer: A. (1)

Explanation: Multiplying both sides by (\sqrt{m}+\sqrt{n}) gives (1=m-n). In exams, apply the conjugate product directly.

Correct answer: B. (2)

Explanation: The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus \(c=1\), \(r=-8\), \(s=7\), and \(c+r+s=0\).