Operations on real numbers and the laws of exponents

The product of denominators is (10-9=1), and the numerator is ((\sqrt{10}+3)-(\sqrt{10}-3)=6). In exams, find the product of conjugate denominators first.

Using the laws of exponents, \(x^{12}=(x^4)^3=5^3=125\) and \(x^8=(x^4)^2=5^2=25\). Therefore, \(x^{12}-x^8=125-25=100\), so option C is correct. Option D is only the value of \(x^{12}\), not of the complete expression. Exam tip: When the required exponents are multiples of a given exponent, rewrite them as powers of that given expression.

Since \(\left(\frac{16}{81}\right)^{\frac{1}{4}}=\frac{2}{3}\), \(\left(\frac{16}{81}\right)^{-\frac{3}{4}}=\left(\frac{2}{3}\right)^{-3}=\frac{27}{8}\). In exams, take the fourth root first.

Here ((3\sqrt{5})^{2}=45), ((2\sqrt{7})^{2}=28), and the middle term is (12\sqrt{35}). Therefore, the expansion is (73-12\sqrt{35}).

Use the law of exponents by expressing both equations with a common base. Since 32 = 2^5, the equation 2^a = 2^5 gives a = 5. Also, 8 = 2^3 and 64 = 2^6, so 8^b = (2^3)^b = 2^(3b) = 2^6; hence 3b = 6 and b = 2. Substituting these values, a^b - b^a = 5^2 - 2^5 = 25 - 32 = -7. Therefore option A is correct. Option B has the opposite sign, while 9 and 13 do not follow from the required exponent calculation.

We have \((4x^{-1})^2=16x^{-2}\) and \((3x^3)^2=9x^6\). Thus, the numerator becomes \(16x^{-2}\cdot9x^6=144x^4\). Therefore, \(\frac{144x^4}{12x^4}=12\), since \(x^4\) cancels for \(x\ne0\). Options B and C retain unnecessary powers of \(x\), while option D is only the numerator. Exam tip: when multiplying like bases, add their exponents; when dividing, subtract them.

Rewrite every numerical factor as a power of 10, then apply the laws of exponents. We have 1000 = 10^3, so 1000^2 = (10^3)^2 = 10^6. Also, 100 = 10^2. Therefore the left-hand side becomes (10^k × 10^6) / 10^2 = 10^(k+6-2) = 10^(k+4). Since it equals 10^9 and the bases are the same, their exponents must be equal: k + 4 = 9. Thus k = 5, so option C is correct. The other values result from mishandling the square on 1000 or the division by 100.

Here (\sqrt{108}=6\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{12}=2\sqrt{3}). The numerator is (9\sqrt{3}), so the value is (9).

The conjugate of \(5+2\sqrt{6}\) is \(5-2\sqrt{6}\). Since \((5+2\sqrt{6})(5-2\sqrt{6})=25-24=1\), we get \(\frac{1}{5+2\sqrt{6}}=5-2\sqrt{6}\). Therefore, \(y+\frac{1}{y}=(5+2\sqrt{6})+(5-2\sqrt{6})=10\). Option A considers only the integer part of \(y\), while option C may result from incorrectly combining the radical terms. Exam tip: use the conjugate to rationalize the denominator in such expressions.

Inside, \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), so its reciprocal is \(x^{2}y^{-4}z^{-3}\). Multiplying by \(\frac{y^{2}}{x^{3}z^{2}}\) gives \(\frac{1}{xy^{2}z^{5}}\).

Here \(64^{\frac{2}{3}}=(4)^{2}=16\) and \(8^{-\frac{4}{3}}=(2)^{-4}=\frac{1}{16}\). The product is (1).

Using the exponent rule \((ab)^k=a^k b^k\), the left-hand side becomes \(x^{-3k}y^{2k}\). Comparing the exponents of the same bases gives \(-3k=-12\) and \(2k=8\), and both equations yield \(k=4\). Exam tip: When monomials with the same bases are equal, compare the corresponding exponents.

We have (\sqrt[3]{216}=6), (\sqrt[3]{a^{12}}=a^{4}), and (\sqrt[3]{b^{9}}=b^{3}). In exams, divide exponents by (3) under a cube root.

Here (3^{-2}+3^{-4}=\frac{1}{9}+\frac{1}{81}=\frac{10}{81}), and (3^{-3}=\frac{1}{27}). Division gives (\frac{10}{3}).

Squaring the given expression, \(x^{2}=(\sqrt{7}-\sqrt{3})^{2}=7+3-2\sqrt{21}=10-2\sqrt{21}\). Therefore, \(x^{2}+2\sqrt{21}=10\), so option C is correct. Exam tip: While using \((a-b)^2=a^2+b^2-2ab\), take care to retain the negative sign in the middle term.

Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).

Express all quantities with the common base 3. Since 729 = 3^6 and 9 = 3^2, the first equation becomes (3^2)^x = 3^(2x) = 3^6, so 2x = 6 and x = 3. Since 27 = 3^3, the second equation becomes (3^3)^y = 3^(3y) = 3^6, giving 3y = 6 and y = 2. Consequently, x + y = 3 + 2 = 5. Therefore option C is the correct answer. Option A is the value of x+y if the bases are mishandled, while 9/2 and 6 do not satisfy the two exponent equations.

Use the conjugate-product identity \((a+b)(a-b)=a^2-b^2\). Thus, \((\sqrt{17}+\sqrt{8})(\sqrt{17}-\sqrt{8})=17-8=9\). Also, \(\sqrt{81}=9\), so the complete expression is \(9-9=0\). Exam tip: Identify conjugate pairs first and apply the difference-of-squares identity instead of expanding the radicals term by term.

Since \(18=2\cdot3^{2}\), we get \(18^{3}=2^{3}\cdot3^{6}\). Therefore, \(\frac{18^{3}}{2^{2}\cdot3^{5}}=\frac{2^{3}\cdot3^{6}}{2^{2}\cdot3^{5}}=2^{3-2}\cdot3^{6-5}=2\cdot3=6\). Hence, the correct answer is 6. Exam tip: when dividing powers with the same base, subtract the exponents using \(a^{m}\div a^{n}=a^{m-n}\).

Here (\frac{1}{s}=\sqrt{10}-3), so (s-\frac{1}{s}=6) and (s+\frac{1}{s}=2\sqrt{10}). Thus (s^{2}-\frac{1}{s^{2}}=12\sqrt{10}).