Operations on real numbers and the laws of exponents

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.

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.

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.