Operations on real numbers and the laws of exponents

Inside, \(\frac{3x^{-2}}{x^{3}}=3x^{-5}\), so \(\left(3x^{-5}\right)^{-2}\cdot x^{-1}=\frac{x^{10}}{9}\cdot x^{-1}=\frac{x^{9}}{9}\). In exams, simplify the bracket first.

The total exponent is ((p+4)+(2p-1)-(p+5)=2p-2), so (2p-2=8) and (p=5). In exams, watch signs while adding and subtracting exponents.

Writing all terms with base (2), the exponent is (7-6+12-8=5). In exams, first convert composite bases into prime bases.

Here (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) and (uv=4). Therefore, the value is (2\sqrt{21}).

The product of denominators is ((3-\sqrt{8})(3+\sqrt{8})=1), and the numerator becomes (2\sqrt{8}). In exams, quickly use the product of conjugate denominators.

Here (5^{x+2}-5^{x}=25\cdot5^{x}-5^{x}=24\cdot5^{x}=600), so (5^{x}=25=5^{2}). In exams, factor out the common power.

Inside, \(a^{-3-2}b^{2-(-4)}=a^{-5}b^{6}\), so raising to (-2) gives \(a^{10}b^{-12}\). In exams, a negative outer power changes the signs of both exponents.

(m^{2}=17+2\sqrt{66}), and the given relation helps compare conjugate forms. Therefore, the intended simplified choice is (34+4\sqrt{66}).

Since \(\left(\frac{125}{216}\right)^{\frac{1}{3}}=\frac{5}{6}\), \(\left(\frac{125}{216}\right)^{-\frac{2}{3}}=\left(\frac{5}{6}\right)^{-2}=\frac{36}{25}\). In exams, take the cube root first and then apply the negative power.

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 \(4x^{4}\).

The direct answer is option B, 4. The powers have the same base, so take the smallest common factor, \(2^x\). Since \(2^{x+1}=2\cdot2^x\) and \(2^{x+2}=4\cdot2^x\), the equation becomes \(2^x+2\cdot2^x+4\cdot2^x=112\). Thus \(7\cdot2^x=112\), so \(2^x=16=2^4\), giving \(x=4\). Option A, 3, would give \(2^x=8\), and the total would be \(56\), not 112. Option B works because it gives \(16+32+64=112\). Option C, 5, gives \(32+64+128=224\). Option D, 6, gives an even larger total, 448. The key is to factor the smallest power before solving. Memory cue: consecutive powers of 2 have multipliers 1, 2, and 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).