Operations on real numbers and the laws of exponents

Since \(\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2\), we get \(25=x^{2}+\frac{1}{x^{2}}+2\). In exams, use the identity and subtract (2).

Here (49^{-1}=7^{-2}) and (343^{-2}=7^{-6}), so (\frac{7^{4}\cdot7^{-2}}{7^{-6}}=7^{8}). In exams, dividing by a negative power adds the exponent.

We have (\sqrt{98}=7\sqrt{2}), (\sqrt{72}=6\sqrt{2}), (\sqrt{32}=4\sqrt{2}), and (\sqrt{18}=3\sqrt{2}), so the value is (2\sqrt{2}). In exams, combine only like radicals.

Since (9^{x-1}=3^{2x-2}), the total exponent is (x+2x-2=3x-2). From (243=3^{5}), (3x-2=5), so (x=\frac{7}{3}).

The numerator is (\frac{y^{2}-x^{2}}{x^{2}y^{2}}) and the denominator is (\frac{y-x}{xy}), so division gives (\frac{x+y}{xy}). In exams, convert negative powers to fractions.

Because ((2+\sqrt{5})^{2}=4+5+4\sqrt{5}=9+4\sqrt{5}), (\sqrt{A}=2+\sqrt{5}). In exams, recognize a perfect-square surd form.

Inside, \(\frac{4x^{3}y^{-2}}{2x^{-1}y^{4}}=2x^{4}y^{-6}\), and its square is \(4x^{8}y^{-12}\). Multiplying by \(\frac{y^{12}}{x^{4}}\) gives \(4x^{4}\).

Factoring (2^{x}), we get (2^{x}(1+2+4)=112), so (7\cdot2^{x}=112). Thus (2^{x}=16) and (x=4).

Here \(27^{\frac{2}{3}}=9\), so the first factor is \(\frac{1}{9}\), and \(81^{\frac{3}{4}}=27\). The product is (3).

Since (r^{2}=10+2+2\sqrt{20}=12+4\sqrt{5}), (r^{2}-4\sqrt{5}=12). In exams, subtract the radical middle term correctly.

Since (x^{4}-16=(x^{2}-4)(x^{2}+4)), cancelling the common factor leaves (x^{2}+4). In exams, recognize the difference of squares.

The product of denominators is (6-5=1), and the numerator is ((\sqrt{6}+\sqrt{5})+(\sqrt{6}-\sqrt{5})=2\sqrt{6}). In exams, adding conjugate fractions is often easier together.

Here (x^{9}=(x^{3})^{3}=8) and (x^{6}=(x^{3})^{2}=4), so the sum is (12). In exams, express powers as multiples of the given power.

Since \(\left(\frac{9}{16}\right)^{\frac{1}{2}}=\frac{3}{4}\), \(\left(\frac{9}{16}\right)^{-\frac{3}{2}}=\left(\frac{3}{4}\right)^{-3}=\frac{64}{27}\). In exams, take the square root, cube, and invert.

Here ((2\sqrt{3})^{2}=12), ((3\sqrt{2})^{2}=18), and the middle term is (2\cdot2\sqrt{3}\cdot3\sqrt{2}=12\sqrt{6}). Therefore, the answer is (30-12\sqrt{6}).

From (2^{a}=2^{4}), (a=4), and from (4^{b}=4^{3}), (b=3). Thus (a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17), so the listed magnitude is (17).

The numerator is ((3x^{2})^{3}(2x^{-1})^{2}=27x^{6}\cdot4x^{-2}=108x^{4}). Then (\frac{108x^{4}}{6x^{4}}=18), so check cancellation of powers.

The governing concept is expressing all factors with the same base before comparing exponents. Since 100=10^2, we have 100^2=(10^2)^2=10^4. Also, 1000=10^3. Hence the left-hand side becomes 10^m · 10^4 / 10^3 = 10^(m+4-3)=10^(m+1). Because this equals 10^6, the exponents must be equal, so m+1=6 and m=5. Option C is therefore correct. The other choices result from failing to square 100, mishandling the denominator, or simply choosing the final exponent without accounting for the extra powers.

Here (\sqrt{75}=5\sqrt{3}) and (\sqrt{48}=4\sqrt{3}), so the numerator is (9\sqrt{3}). Dividing by (\sqrt{3}) gives (9).

Since (\frac{1}{3+2\sqrt{2}}=3-2\sqrt{2}), the sum is (6). In exams, use the conjugate quickly for such reciprocals.