Operations on real numbers and the laws of exponents

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

Explanation: A negative exponent represents a reciprocal. Thus, \((ab)^{-1}=\frac{1}{ab}=\frac{1}{a}\cdot\frac{1}{b}=a^{-1}b^{-1}\). No similar rule applies to a sum or difference. Exam tip: distinguish a product from a sum before using exponent laws.

Correct answer: A. \(x^3y\)

Explanation: First, \((3x^2y^{-1})^2=9x^4y^{-2}\). Hence, \(\frac{9x^4y^{-2}}{9xy^{-3}}=x^{4-1}y^{-2-(-3)}=x^3y\). Therefore, option A is correct. Exam tip: when dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

Correct answer: A. \(5^5\)

Explanation: When powers with the same base are multiplied, their exponents are added: \(5^4\cdot5^{-2}\cdot5^3=5^{4+(-2)+3}=5^5\). Therefore, option A is correct. Option B incorrectly treats the negative exponent as positive, while option C results from an incomplete combination of the exponents. Exam tip: for the same base, use \(a^m\cdot a^n=a^{m+n}\), keeping the sign of every exponent.

Correct answer: A. \(7^2\)

Explanation: For multiplication of powers with the same base, add the exponents: \(7^{-1}\cdot 7^2=7^{-1+2}=7^1\). Then divide powers with the same base by subtracting exponents: \(\frac{7^3}{7^1}=7^{3-1}=7^2\). Hence, option A is correct. Exam tip: add exponents when multiplying like bases and subtract them when dividing like bases.

Correct answer: A. \(x^3\)

Explanation: Using the laws of exponents, \((x^2)^3=x^6\). Hence, \(\frac{x^6\cdot x^{-1}}{x^2}=x^{6-1-2}=x^3\). Therefore, option A is correct. Option B results from not correctly subtracting the exponent of the denominator. Exam tip: add exponents when multiplying powers with the same base and subtract them when dividing.

Correct answer: A. \((a^m)^n=a^{mn}\)

Explanation: For a power raised to another power, the exponents are multiplied: \((a^m)^n=a^{mn}\). For example, \((x^2)^3=x^{2\times3}=x^6\). The rule \(m+n\) is used when powers with the same base are multiplied. Exam tip: identify nested powers and multiply their exponents.

Correct answer: B. \((ab)^r=a^r b^r\)

Explanation: The power-of-a-product law is \((ab)^r=a^r b^r\): the exponent applies to each factor. Option A wrongly changes multiplication into addition. Exam tip: an exponent outside brackets acts on the entire product, not on a sum.

Correct answer: C. 16

Explanation: Since (0.25)=\frac{1}{4}, we get (0.25)^{-2}=\left(\frac{1}{4}\right)^{-2}=4^2=16. A negative exponent means taking the reciprocal before applying the positive exponent. Exam tip: use a^{-n}=\frac{1}{a^n}; the option 0.0625 results from squaring 0.25 without handling the negative exponent.

Correct answer: A. (1)

Explanation: \(\left(\frac{4}{3}\right)^{-2}=\left(\frac{3}{4}\right)^2=\frac{9}{16}\). Hence \(\frac{9}{16}\cdot\frac{16}{9}=1\).

Correct answer: B. \(\frac{28}{3}\)

Explanation: Substituting \(a=3\), we get \(a^2=3^2=9\) and \(a^{-1}=\frac{1}{a}=\frac{1}{3}\). Therefore, \(a^2+a^{-1}=9+\frac{1}{3}=\frac{28}{3}\). Option A results from confusing the addition with subtraction; remember that a negative exponent denotes a reciprocal, not a negative value.

Correct answer: A. \(a^9b^6\)

Explanation: Using the exponent law \((xy)^n=x^ny^n\), the outer exponent 3 multiplies the exponent of each factor: \((a^3b^2)^3=a^{3\times3}b^{2\times3}=a^9b^6\). Therefore, option A is correct. In option C, only the exponent of \(a\) is multiplied, while the exponent of \(b\) is incorrectly left as 2. Exam tip: when a power applies to a product, distribute it to every factor and multiply the exponents.

Correct answer: A. \(x^3\)

Explanation: Using the laws of exponents, \((3x^2)^2=3^2x^4=9x^4\). Therefore, \(\frac{9x^4}{9x}=x^3\), since division by 9x is valid when \(x\neq0\). Options B and D incorrectly retain an extra coefficient of 3 or 9. Exam tip: apply the power to every factor in the bracket before cancelling common factors.

Correct answer: A. \(a^{-n}=\frac{1}{a^n}\)

Explanation: A negative exponent represents a reciprocal, so \(a^{-n}=\frac{1}{a^n}\). Check: \(a^n\cdot a^{-n}=a^0=1\). Here \(a\neq0\) is essential. Exam tip: rewrite negative powers as reciprocals first.

Correct answer: B. 2

Explanation: Substituting x = −1 gives q(−1) = 2(−1)³ + (−1)² − 3(−1) = −2 + 1 + 3 = 2. Therefore, option B is correct. Exam tip: always use parentheses when substituting a negative number to avoid sign errors, especially in terms such as −3(−1).

Correct answer: A. \\(6x^3\\)

Explanation: All the terms are like terms because they have the same variable \(x\) with the same exponent 3. Therefore, only their coefficients are combined: \(4+7-5=6\). Hence, the simplified form is \(6x^3\). Do not add the exponents here; exponents are added when powers with the same base are multiplied.

Correct answer: A. 6x - 4

Explanation: Combine the like terms: 2x + 4x = 6x, and combine the constants: 3 + (-7) = -4. Therefore, the simplified form is 6x - 4. Option B incorrectly treats the constant terms as 3 + 7 = 10. In exams, group variable terms and constant terms separately before adding.

Correct answer: A. \(4x-7\)

Explanation: Distribute the minus sign before the second bracket to both of its terms: \\(7x-2-3x-5\\). Combining like terms gives \\(7x-3x-2-5=4x-7\\), so option A is correct. Option B results from handling the constant terms incorrectly. Exam tip: when a bracket is preceded by a minus sign, change the sign of every term inside that bracket.

Correct answer: A. \(6x^3-15x^2+12x\)

Explanation: Using the distributive law, multiply \(3x\) by each term inside the parentheses: \(3x\cdot2x^2=6x^3\), \(3x\cdot(-5x)=-15x^2\), and \(3x\cdot4=12x\). Therefore, the expanded form is \(6x^3-15x^2+12x\). Option B fails to multiply each term by \(x\), while option D has the wrong sign for the middle term. Exam tip: when removing parentheses, multiply the outside term by every term inside and use \(x^m\cdot x^n=x^{m+n}\) for powers.

Correct answer: A.
x^2+7x+12

Explanation: Using the distributive property,
(x+4)(x+3)=x^2+3x+4x+12. Combining like terms gives
x^2+7x+12. Option B incorrectly interchanges the coefficient of x and the constant term. Exam tip: use
(x+a)(x+b)=x^2+(a+b)x+ab.

Correct answer: A. \(a^m \times a^n=a^{m+n}\)

Explanation: When powers with the same base are multiplied, their exponents are added, so \(a^m\times a^n=a^{m+n}\) is correct. For example, \(a^2\times a^3=a^5\). In \((a^m)^n\), exponents are multiplied. Exam tip: check whether the bases are the same first.

Correct answer: A. (3x^2-11x-4)

Explanation: Using the distributive law, (3x+1)(x-4)=3x^2-12x+x-4. Combining like terms gives -12x+x=-11x, so the expansion is (3x^2-11x-4). Option B has the wrong sign for the middle term. Exam tip: multiply every term in one binomial by both terms in the other binomial.

Correct answer: A. \\(x^2+16x+64\\)

Explanation: Apply the identity \\(a+b\\)^2=a^2+2ab+b^2\\). With \\(a=x\\) and \\(b=8\\), we get \\(x^2+2(x)(8)+8^2=x^2+16x+64\\). Option B omits the middle term, while option C uses an incorrect coefficient for it. In an exam, always check that the middle term is \\(2ab\\).

Correct answer: A. \\(x^2-18x+81\\)

Explanation: Use the identity \\((a-b)^2=a^2-2ab+b^2\\). Here, \\(a=x\\) and \\(b=9\\), so \\(x^2-2(x)(9)+9^2=x^2-18x+81\\). Option B has the wrong sign in the middle term because this is the square of a difference, not a sum. In an exam, remember that the middle term in \\((a-b)^2\\) is always \\(-2ab\\).

Correct answer: A. \(a^m \div a^n=a^{m-n}\)

Explanation: When powers with the same base are divided, their exponents are subtracted: \(a^m\div a^n=a^{m-n}\), where \(a\ne0\). Option B is the multiplication rule, not the quotient rule. In exams, first check that the bases match and the base is non-zero.

Correct answer: A. \((8x-9)(8x+9)\)

Explanation: \(64x^2-81=(8x)^2-9^2\). Using the difference of squares identity \(a^2-b^2=(a-b)(a+b)\), we get \((8x-9)(8x+9)\). Options C and D are squares of binomials, which would contain a middle term, so they are not correct. Exam tip: Whenever two perfect squares are being subtracted, check for the difference of squares identity first.