Operations on real numbers and the laws of exponents

Correct answer: A. (1)

Explanation: Here (pq=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=3-2=1). In exams, use ((a+b)(a-b)=a^{2}-b^{2}).

Correct answer: A. (2+\sqrt{3})

Explanation: To rationalize, (\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}). In exams, multiply by the conjugate.

Correct answer: A. (\frac{9}{8})

Explanation: Here (x=\frac{8}{9}), so (x^{-1}=\frac{9}{8}). In exams, apply (a^{-n}=\frac{1}{a^{n}}) in the correct direction.

Correct answer: A. \(\frac{y^{8}}{4x^{6}}\)

Explanation: Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.

Correct answer: B. (4)

Explanation: Here (2^{x+1}+2^{x}=2\cdot2^{x}+2^{x}=3\cdot2^{x}=48), so (2^{x}=16=2^{4}). In exams, factor the common power (2^{x}).

Correct answer: A. (6\sqrt{2})

Explanation: We get (\sqrt{50}=5\sqrt{2}), (\sqrt{18}=3\sqrt{2}), and (\sqrt{8}=2\sqrt{2}), so the result is (6\sqrt{2}). In exams, combine only like surd terms.

Correct answer: A. \(\frac{4}{9}\)

Explanation: Since \(\left(\frac{27}{8}\right)^{\frac{1}{3}}=\frac{3}{2}\), \(\left(\frac{27}{8}\right)^{-\frac{2}{3}}=\left(\frac{3}{2}\right)^{-2}=\frac{4}{9}\). In exams, take the cube root first.

Correct answer: A. (\sqrt{5}-2)

Explanation: Rationalizing gives (\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2). In exams, use the conjugate of the denominator.

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

Explanation: Here \(\left(9^{2}\right)^{3}=(3^{2})^{6}=3^{12}\), and \(3^{12}\div3^{10}=3^{2}\). In exams, write (9) as \(3^{2}\).

Correct answer: A. \(a^{2}b^{-1}\)

Explanation: We have \(\left(ab^{-2}\right)^{3}=a^{3}b^{-6}\), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.

Correct answer: A. (12\sqrt{2})

Explanation: Using (x^{2}-y^{2}=(x-y)(x+y)), we get (x-y=2\sqrt{2}) and (x+y=6), so the value is (12\sqrt{2}). In exams, identities reduce calculation.

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

Explanation: Here ((2^{5})^{3}=2^{15}), (4^{-2}=2^{-4}), and (8^{2}=2^{6}), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).

Correct answer: A. (x)

Explanation: Since (\sqrt{x^{3}}=x^{\frac{3}{2}}), (x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x). In exams, convert radicals to fractional exponents.

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

Explanation: Inside, \(\frac{x^{-2}y^{3}}{x^{4}y^{-1}}=x^{-6}y^{4}\), and raising to (-1) gives \(x^{6}y^{-4}\). In exams, subtract exponents during division.

Correct answer: A. (14)

Explanation: Here \(\frac{1}{a}=2-\sqrt{3}\), so \(a+\frac{1}{a}=4\) and \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\). In exams, use the identity \(\left(a+\frac{1}{a}\right)^{2}\).

Correct answer: A. \(2+2\sqrt{5}\)

Explanation: The first product is \(7-5=2\), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.

Correct answer: C. (8)

Explanation: Since (125=5^{3}), (x=3), and since (32=2^{5}), (y=5), so (x+y=8). In exams, write numbers as powers of their prime bases.

Correct answer: A. (12)

Explanation: Here (3^{-2}+3^{-1}=\frac{1}{9}+\frac{1}{3}=\frac{4}{9}), and (3^{-3}=\frac{1}{27}), so the value is (12). In exams, convert negative powers into fractions.

Correct answer: A. \((x-2)(x+2)\)

Explanation: We use \(x^{2}-4=x^{2}-2^{2}=(x-2)(x+2)\). In exams, remember the difference of squares form \((a^{2}-b^{2})\).

Correct answer: A. (8+4\sqrt{3})

Explanation: ((\sqrt{6}+\sqrt{2})^{2}=6+2+2\sqrt{12}=8+4\sqrt{3}). In exams, do not miss the middle term of ((a+b)^{2}).

Correct answer: A. (10)

Explanation: For the same base (10), the exponent is (2-5+4=1), so (r=10^{1}=10). In exams, add exponents during multiplication.

Correct answer: A. \(a^{8}b^{-8}\)

Explanation: Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.

Correct answer: A. (4)

Explanation: (\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\frac{1}{2}), so (u^{-2}=4). In exams, simplify the radical first.

Correct answer: A. \(a^{0}=1\)

Explanation: The zero-exponent law states that \(a^{0}=1\) for every nonzero number, so option A is correct. Option B is false because \(a^{0}\) equals 1, not 0. In option C, \(a^{-1}=\frac{1}{a}\), which is not generally equal to \(a\); and in option D, \(a^{1}=a\). Exam tip: Apply \(a^{0}=1\) only when \(a\ne0\); do not include \(0^{0}\) in this rule.

Correct answer: A. (4)

Explanation: Since (\frac{1}{2-\sqrt{3}}=2+\sqrt{3}), (\frac{1}{x}+x=(2+\sqrt{3})+(2-\sqrt{3})=4). In exams, identify conjugate numbers quickly.