Operations on real numbers and the laws of exponents

Given \(a=2\), we have \(a^3=2^3=8\). By the negative exponent rule, \(a^{-1}=\frac{1}{a}=\frac{1}{2}\). Therefore, \(a^3+a^{-1}=8+\frac{1}{2}=\frac{17}{2}\). The value \(\frac{15}{2}\) results from evaluating one of the powers incorrectly. Exam tip: remember that \(x^{-n}=\frac{1}{x^n}\) for \(x\neq0\).

Using the exponent rule \\( (ab)^n=a^n b^n \\), we get \\( (x^2y^3)^2=(x^2)^2(y^3)^2=x^4y^6 \\). Therefore, option A is correct. In option B, the exponent of \\(y\\) has been handled incorrectly; the outside exponent must multiply 3 to give 6. Exam tip: when a product is raised to a power, apply that power to every factor.

First apply the power rule: \((2x^3)^2=2^2x^{3\times2}=4x^6\). Then \(\frac{4x^6}{4x^2}=x^{6-2}=x^4\), so option A is correct. Option D, \(x^6\), results from failing to subtract the exponent in the denominator. Exam tip: when dividing powers with the same non-zero base, subtract the exponents.

Substitute \(x=2\) into the polynomial: \(p(2)=2(2)^2-3(2)+1=2\times4-6+1=3\). Therefore, the correct value is 3. Exam tip: evaluate the power first, followed by multiplication and then addition or subtraction.

Substitute \(x=-1\) into the polynomial: \(q(-1)=(-1)^3-2(-1)^2+(-1)=-1-2-1=-4\). Thus, the correct answer is -4. Since \((-1)^2=1\), the middle term is \(-2\), not \(+2\). In exams, always use parentheses when substituting a negative value.

All the terms have the same variable and the same exponent, so they are like terms. Combine their coefficients: 3 + 5 - 2 = 6. Therefore, the simplified form is 6x^2. Option C is incorrect because it treats the negative term as positive. Exam tip: When combining like terms, operate only on the coefficients; the variable and its exponent remain unchanged.

When powers with the same non-zero base are divided, their exponents are subtracted: \(\dfrac{a^m}{a^n}=a^{m-n}\). Hence, \(\dfrac{a^7}{a^3}=a^{7-3}=a^4\). Option B incorrectly adds the exponents, a rule used for multiplication of powers with the same base. Exam tip: for division, subtract the denominator’s exponent from the numerator’s exponent.

When powers have the same base, their exponents are added, so \(a^m\times a^n=a^{m+n}\). The exponent \(m-n\) is used for division, not multiplication. Exam tip: first check whether the bases are identical.

By the distributive law, multiply 2x by each term inside the bracket: 2x·3x^2=6x^3, 2x·(-4x)=-8x^2, and 2x·1=2x. Therefore, the expanded form is 6x^3-8x^2+2x. Option B does not correctly account for multiplication by 2x, while option D has the wrong sign for the middle term. Exam tip: Multiply the term outside the bracket by every term inside it, including the constant term.

Using the distributive law, multiply each term: x·x + x·2 + 3·x + 3·2 = x² + 2x + 3x + 6. Combining the like terms 2x and 3x gives x² + 5x + 6. Option B has the coefficients of the middle and constant terms incorrect. Exam tip: for two binomials, multiply the first, outer, inner and last terms, then combine like terms.

Use the power-of-a-power law \((a^m)^n=a^{mn}\) and the quotient law for equal bases, \(a^m\div a^n=a^{m-n}\). Thus, \(\frac{(a^4)^3}{a^5}=\frac{a^{12}}{a^5}=a^{12-5}=a^7\). Therefore, option B is correct. Option C stops after applying the power-of-a-power rule, while option D incorrectly adds the exponents during division. In an exam, multiply exponents inside a power first and subtract exponents when dividing like bases.

Using the distributive property, multiply both \(2x\) and \(3\) by each term in \(x-2\): \((2x+3)(x-2)=2x^2-4x+3x-6\). Combining like terms, \(-4x+3x=-x\), gives \(2x^2-x-6\). Hence, option A is correct. Option B has an incorrect positive sign for the middle term. Exam tip: write all four products before combining like terms.

When powers with the same non-zero base are divided, their exponents are subtracted: \(a^m/a^n=a^{m-n}\). For example, \(a^7/a^3=a^4\). The \(m+n\) rule applies to multiplication. In exams, check that the base is non-zero.

Use the identity (a-b)^2=a^2-2ab+b^2. Substituting a=x and b=6 gives (x-6)^2=x^2-2(x)(6)+6^2=x^2-12x+36, so option A is correct. Option B has the wrong sign for the middle term. In exams, remember that the middle term of (a-b)^2 is -2ab.

The expression uses the difference-of-squares identity \((a+b)(a-b)=a²-b²\). Here, \(a=x
u0000\) and \(b=7
u0000\), so \((x+7)(x-7)=x²-7²=x²-49
u0000\). Option B has the wrong sign, while options C and D do not result from multiplying these conjugate binomials. In an exam, identify \((a+b)(a-b)\) directly as \(a²-b²\).

\(49x^2-64=(7x)^2-8^2\). Using the difference of squares identity \(a^2-b^2=(a-b)(a+b)\), we get \((7x-8)(7x+8)\). Option B does not expand to the given expression, while C and D are squares rather than the correct difference-of-squares factorization. Exam tip: first rewrite both terms as perfect squares.

Use the perfect-square identity \((a+b)^2=a^2+2ab+b^2\). Taking \(a=x\) and \(b=7\), we get \((x+7)^2=x^2+2\cdot x\cdot7+7^2=x^2+14x+49\), so option A is correct. Option B would produce the middle term \(-14x\), not \(+14x\). Exam tip: take the square root of the constant term and use the sign of the middle term to choose the appropriate binomial.

The expression can be written as \(x^2-2\cdot x\cdot 8+8^2\). Using \(a^2-2ab+b^2=(a-b)^2\), it equals \((x-8)^2\). Option B would produce a middle term of \(+16x\), so it is incorrect. Exam tip: identify the square roots of the first and last terms, then check whether the middle term is \(-2ab\) or \(+2ab\).

Use the identity \(m^2+n^2=(m+n)^2-2mn\). Thus, \(m^2+n^2=9^2-2(18)=81-36=45\), so the correct answer is 45. The option 63 may result from an incorrect calculation while subtracting \(2mn\). Exam tip: remember the identity \(a^2+b^2=(a+b)^2-2ab\) for such questions.

Use the identity \(a^2+b^2=(a-b)^2+2ab\). Thus, \(a^2+b^2=5^2+2(6)=25+12=37\). Option B results from incorrectly adding only \(ab\) instead of \(2ab\). In exams, remember that \((a-b)^2=a^2-2ab+b^2\), which leads to this identity.