Operations on real numbers and the laws of exponents

First, \(\frac{x^{-3}}{y^{-2}}=x^{-3}\times y^2=\frac{y^2}{x^3}\). Applying the outer power \(-1\) takes the reciprocal: \(\left(\frac{y^2}{x^3}\right)^{-1}=\frac{x^3}{y^2}\). Therefore, option A is correct. Option B is only the simplified inner expression and does not include the effect of the outer \(-1\) power. Exam tip: the \(-1\) power of any nonzero expression gives its reciprocal.

Any non-zero number raised to the power 0 equals 1, so \((3^2)^0=1\). Using the negative-exponent rule \(a^{-n}=\frac{1}{a^n}\), we get \(4^{-1}=\frac{1}{4}\). Therefore, \(1+\frac{1}{4}=\frac{5}{4}\), so option B is correct. Option A incorrectly omits the contribution of the first term. Exam tip: apply the zero-exponent and negative-exponent rules separately before adding.

Using the negative exponent rule \(a^{-n}=\frac{1}{a^n}\), we get \(9^{-1}=\frac{1}{9}\) and \(3^{-2}=\frac{1}{3^2}=\frac{1}{9}\). Therefore, the sum is \(\frac{1}{9}+\frac{1}{9}=\frac{2}{9}\). A useful exam tip is to convert negative powers into reciprocals before performing the operation.

Since \(9=3^2\), we have \(9^{-1}=(3^2)^{-1}=3^{-2}\). Therefore, \(3^5\cdot 9^{-1}=3^5\cdot 3^{-2}=3^{5-2}=3^3=27\), so option B is correct. Exam tip: When multiplying powers with the same base, add their exponents.

Apply the identity \((a-b)^2=a^2-2ab+b^2\). Here, \(a=3x\) and \(b=4\). Thus, \((3x-4)^2=(3x)^2-2(3x)(4)+4^2=9x^2-24x+16\). Therefore, option A is correct. Remember that the middle term is \(-2ab\), not just \(-ab\).

For real exponents, \(a^m\) is generally defined for \(a>0\), so the product law applies safely. With \(a<0\), some irrational powers are not real. Exam tip: whenever real exponents appear, first check that the base is positive.

Use the identity \((a+b)(a-b)=a^2-b^2\). Here, \(a=6x\) and \(b=5\), so \((6x)^2-5^2=36x^2-25\). Therefore, option A is correct. Option B incorrectly uses the sum of the squares, while the given factors form a difference of squares. Exam tip: whenever two binomials differ only in the sign between the terms, check the identity \((a+b)(a-b)\).

Since \(64=2^6\), the equation becomes \(2^{x+1}=2^6\), so \(x+1=6\). Therefore, \(x=5\). Choosing 6 is a common mistake because it is the value of \(x+1\), not \(x\). In the exam, first express both sides with the same base and then equate the exponents.

Since \(125=5^3\), the equation becomes \(5^{x-2}=5^3\). With equal bases, the exponents are equal, so \(x-2=3\), giving \(x=5\). Exam tip: rewrite the numerical side as a power of the same base before equating exponents.

The governing concept is the law of exponents: \((a^m)^n=a^{mn}\), negative exponents represent reciprocals, and division by a power subtracts its exponent. Rewrite every quantity with base 2. We have \((2^3)^2=2^6\), \(4^{-1}=(2^2)^{-1}=2^{-2}\), and \(8=2^3\). Therefore the expression is \(2^6\cdot2^{-2}\div2^3=2^{6-2-3}=2^1=2\). Hence option A is correct. A common error is to treat \(4^{-1}\) as 4 or to add the denominator exponent instead of subtracting it. Options B, C and D reflect such exponent or reciprocal mistakes.

Here, \(81x^2=(9x)^2\) and \(100y^2=(10y)^2\). Thus, the expression becomes \((9x)^2-(10y)^2\). Using the difference of squares identity \(a^2-b^2=(a-b)(a+b)\), we get \((9x-10y)(9x+10y)\). Options C and D are squares whose expansions introduce an additional \(90xy\) term, so they are incorrect. Exam tip: in the difference of two squares, the two factors differ only in the sign between the terms.

Using the power-of-a-power rule, \((3^2)^4=3^{2\times4}=3^8\). For multiplication of powers with the same base, add the exponents: \(3^5\cdot3^{-2}=3^{5+(-2)}=3^3\). Therefore, \(\frac{3^8}{3^3}=3^{8-3}=3^5\), so option A is correct. Option B gives only the denominator’s power, not the simplified value of the whole fraction. Exam tip: add exponents when multiplying like bases and subtract them when dividing.

Rewrite every quantity using base 2. We have \(4^3=(2^2)^3=2^6\u0005, \(2^{-1}\u0005 remains as it is, and \(8=2^3\u0005. Therefore the expression becomes \(\frac{2^6\cdot2^{-1}}{2^3}\u0005. Using exponent laws, multiplication adds exponents and division subtracts them, so the total exponent is \(6+(-1)-3=2\u0005.

Thus the value is \(2^2=4\u0005, making option B correct. The negative exponent means reciprocal, since \(2^{-1}=\frac12\u0005; it does not mean that the final answer is negative. Direct calculation also gives \(64\cdot\frac12\div8=4\u0005.

For powers with the same nonzero base, exponents are added during multiplication and subtracted during division. Thus, \(\frac{x^7\cdot x^{-3}}{x^2}=x^{7+(-3)-2}=x^2\). Option C results from simplifying only \(x^7\cdot x^{-3}\) and ignoring the division by \(x^2\). Exam tip: subtract the denominator’s exponent when dividing like bases.

When a quotient is raised to a power, the exponent applies to both numerator and denominator: \(\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}\). In B, the denominator lacks the exponent. Exam tip: apply the power to every factor.

To simplify a quotient of monomials, divide the numerical coefficients and subtract the exponent of each variable in the denominator from its exponent in the numerator. The coefficient becomes \(18\div6=3\u0005. For \(x\u0005, the exponent is \(5-2=3\u0005, and for \(y\u0005, it is \(3-1=2\u0005. The conditions \(x\ne0\u0005 and \(y\ne0\u0005 allow this cancellation.

Therefore the simplified expression is \(3x^3y^2\u0005, so option A is correct. Exponents are subtracted because \(a^m/a^n=a^{m-n}\u0005 for nonzero \(a\u0005. Adding exponents would apply to multiplication, not division, which is why the other forms are incorrect.

\(\left(\frac{1}{5}\right)^{-2}=25\) and \(\left(\frac{1}{3}\right)^{-2}=9\). Therefore the sum is (34).

For every non-zero real number, the zero-exponent law gives \(a^0=1\). In contrast, \(a^{-1}=1/a\), which is not generally 1. Exam tip: always check that the base is non-zero before applying this law.

By the law of exponents, \(a^{-1}=\frac{1}{a}\). Therefore, \(5^2-5^{-1}=25-\frac{1}{5}=\frac{125-1}{5}=\frac{124}{5}\). Option B results from incorrectly adding 1 in the numerator instead of subtracting it. In an exam, remember that a negative exponent represents the reciprocal of the base.

Substitute \(x=3\) into the polynomial: \(p(3)=4(3)^2-5(3)+2=4\times9-15+2=36-15+2=23\). Therefore, the correct value is 23. Exam tip: evaluate the exponent before multiplication, and do not interpret \(4(3)^2\) as \((4\times3)^2\).