Operations on real numbers and the laws of exponents

Correct answer: A. (18)

Explanation: The numerator is ((3x^{2})^{3}(2x^{-1})^{2}=27x^{6}\cdot4x^{-2}=108x^{4}). Then (\frac{108x^{4}}{6x^{4}}=18), so check cancellation of powers.

Correct answer: A. (9)

Explanation: Here (\sqrt{75}=5\sqrt{3}) and (\sqrt{48}=4\sqrt{3}), so the numerator is (9\sqrt{3}). Dividing by (\sqrt{3}) gives (9).

Correct answer: A. (6)

Explanation: Since (\frac{1}{3+2\sqrt{2}}=3-2\sqrt{2}), the sum is (6). In exams, use the conjugate quickly for such reciprocals.

Correct answer: A. \(\frac{y}{xz}\)

Explanation: Inside, \(\frac{x^{3}y^{-2}}{z^{-1}}=x^{3}y^{-2}z\), so its reciprocal is \(x^{-3}y^{2}z^{-1}\). Multiplying by \(\frac{x^{2}}{yz^{2}}\) gives \(\frac{y}{xz^{3}}\), so the (z)-power must be checked carefully.

Correct answer: A. \(\frac{1}{2}\)

Explanation: Here \(32^{\frac{2}{5}}=(2^{5})^{\frac{2}{5}}=2^{2}=4\), and \(4^{-\frac{3}{2}}=(2^{2})^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}\). The product is \(\frac{1}{2}\).

Correct answer: C. 5

Explanation: Using the exponent rule \((a^m)^n=a^{mn}\), we get \((x^{2}y^{-1})^k=x^{2k}y^{-k}\). Comparing the exponents of the same bases gives \(2k=10\) and \(-k=-5\), so both equations yield \(k=5\). Option 4 is incorrect because it would produce \(x^8y^{-4}\). Exam tip: multiply every exponent inside the bracket by the outside exponent before comparing like bases.

Correct answer: A. (5a^{3}b^{2})

Explanation: We have (\sqrt[3]{125}=5), (\sqrt[3]{a^{9}}=a^{3}), and (\sqrt[3]{b^{6}}=b^{2}). In exams, divide exponents by (3) under a cube root.

Correct answer: A. (x^{3}+1)

Explanation: Since (x^{6}-1=(x^{3}-1)(x^{3}+1)), cancelling the common factor gives (x^{3}+1). In exams, recognize the (A^{2}-B^{2}) form.

Correct answer: A. (2\sqrt{15})

Explanation: Since (\frac{1}{4-\sqrt{15}}=4+\sqrt{15}), (\frac{1}{p}-p=(4+\sqrt{15})-(4-\sqrt{15})=2\sqrt{15}). In exams, the conjugate gives the reciprocal directly when the denominator product is (1).

Correct answer: A. (6)

Explanation: Here (2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}), and (2^{-5}=\frac{1}{32}). Therefore, the value is (\frac{3}{16}\div\frac{1}{32}=6).

Correct answer: A. (7)

Explanation: Since (x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}), (x^{2}+2\sqrt{10}=7). In exams, write the middle term of ((a-b)^{2}) carefully.

Correct answer: A. \(5m^{6}n^{-4}\)

Explanation: Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.

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

Explanation: Since (216=6^{3}), (x=3). Also (36^{y}=(6^{2})^{y}=6^{2y}=6^{3}), so (y=\frac{3}{2}) and the sum is (\frac{9}{2}).

Correct answer: A. (0)

Explanation: The conjugate product is \(13-3=10\), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.

Correct answer: A. (2^{3}\cdot3)

Explanation: Since (12^{4}=(2^{2}\cdot3)^{4}=2^{8}\cdot3^{4}), division leaves (2^{3}\cdot3). In exams, prime-factorize first.

Correct answer: A. (8\sqrt{7})

Explanation: Here (\frac{1}{s}=\sqrt{7}-2), so (s-\frac{1}{s}=4) and (s+\frac{1}{s}=2\sqrt{7}). Thus (s^{2}-\frac{1}{s^{2}}=8\sqrt{7}).

Correct answer: A. \(\frac{2xy^{2}}{3}\)

Explanation: We get \(\left(\frac{27x^{-3}}{8y^{6}}\right)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}\), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.

Correct answer: A. (6)

Explanation: Since \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\), \(24=4\left(x+\frac{1}{x}\right)\). In exams, use the difference of squares identity.

Correct answer: A. \(a\)

Explanation: Combining like terms gives \(2a^{-1}+3a^{-1}=5a^{-1}\). Therefore, \(\frac{5a^{-1}}{5a^{-2}}=a^{-1-(-2)}=a^1=a\), so option A is correct. Exam tip: when dividing powers with the same non-zero base, subtract the exponents: \(a^m/a^n=a^{m-n}\).

Correct answer: A. (54\sqrt{2})

Explanation: From (\sqrt{x}=3\sqrt{2}), (x=18), and (x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).

Correct answer: A. \(\frac{706}{225}\)

Explanation: Here \(\left(\frac{3}{5}\right)^{-2}=\frac{25}{9}\) and \(\left(\frac{5}{3}\right)^{-2}=\frac{9}{25}\), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.

Correct answer: A. \(\frac{7}{3}\)

Explanation: The left side is \(3^{2x}\cdot3^{x-1}=3^{3x-1}\), and \(729=3^{6}\). Hence \(3x-1=6\) and \(x=\frac{7}{3}\).

Correct answer: A. (16)

Explanation: Here (\sqrt{147}=7\sqrt{3}), (2\sqrt{12}=4\sqrt{3}), and (3\sqrt{27}=9\sqrt{3}), so the numerator is (12\sqrt{3}). Therefore, the value should be (12).

Correct answer: A. (1)

Explanation: Multiplying both sides by (\sqrt{a}+\sqrt{b}), we get (1=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b). In exams, apply the conjugate product directly.

Correct answer: A. \(\frac{x^{6}}{4}\)

Explanation: Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so \(\left(2x^{-5}\right)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}\). In exams, subtract the inner exponents first.