Operations on real numbers and the laws of exponents

Inside, \(\frac{x^{3}y^{-2}}{z^{-1}}=x^{3}y^{-2}z\), so its reciprocal is \(x^{-3}y^{2}z^{-1}\). Multiplying by \(\frac{x^{2}}{yz^{2}}\) gives \(\frac{y}{xz^{3}}\), so the (z)-power must be checked carefully.

Here \(32^{\frac{2}{5}}=(2^{5})^{\frac{2}{5}}=2^{2}=4\), and \(4^{-\frac{3}{2}}=(2^{2})^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}\). The product is \(\frac{1}{2}\).

Using the exponent rule \((a^m)^n=a^{mn}\), we get \((x^{2}y^{-1})^k=x^{2k}y^{-k}\). Comparing the exponents of the same bases gives \(2k=10\) and \(-k=-5\), so both equations yield \(k=5\). Option 4 is incorrect because it would produce \(x^8y^{-4}\). Exam tip: multiply every exponent inside the bracket by the outside exponent before comparing like bases.

We have (\sqrt[3]{125}=5), (\sqrt[3]{a^{9}}=a^{3}), and (\sqrt[3]{b^{6}}=b^{2}). In exams, divide exponents by (3) under a cube root.

Since (x^{6}-1=(x^{3}-1)(x^{3}+1)), cancelling the common factor gives (x^{3}+1). In exams, recognize the (A^{2}-B^{2}) form.

Since (\frac{1}{4-\sqrt{15}}=4+\sqrt{15}), (\frac{1}{p}-p=(4+\sqrt{15})-(4-\sqrt{15})=2\sqrt{15}). In exams, the conjugate gives the reciprocal directly when the denominator product is (1).

Here (2^{-3}+2^{-4}=\frac{1}{8}+\frac{1}{16}=\frac{3}{16}), and (2^{-5}=\frac{1}{32}). Therefore, the value is (\frac{3}{16}\div\frac{1}{32}=6).

Since (x^{2}=5+2-2\sqrt{10}=7-2\sqrt{10}), (x^{2}+2\sqrt{10}=7). In exams, write the middle term of ((a-b)^{2}) carefully.

Inside, \(\frac{5m^{-2}n^{3}}{25m^{4}n^{-1}}=\frac{1}{5}m^{-6}n^{4}\), so raising to (-1) gives \(5m^{6}n^{-4}\). In exams, do not forget to invert the coefficient too.

Since (216=6^{3}), (x=3). Also (36^{y}=(6^{2})^{y}=6^{2y}=6^{3}), so (y=\frac{3}{2}) and the sum is (\frac{9}{2}).

The conjugate product is \(13-3=10\), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.

Since (12^{4}=(2^{2}\cdot3)^{4}=2^{8}\cdot3^{4}), division leaves (2^{3}\cdot3). In exams, prime-factorize first.

Here (\frac{1}{s}=\sqrt{7}-2), so (s-\frac{1}{s}=4) and (s+\frac{1}{s}=2\sqrt{7}). Thus (s^{2}-\frac{1}{s^{2}}=8\sqrt{7}).

We get \(\left(\frac{27x^{-3}}{8y^{6}}\right)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}\), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.

Since \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\), \(24=4\left(x+\frac{1}{x}\right)\). In exams, use the difference of squares identity.

Combining like terms gives \(2a^{-1}+3a^{-1}=5a^{-1}\). Therefore, \(\frac{5a^{-1}}{5a^{-2}}=a^{-1-(-2)}=a^1=a\), so option A is correct. Exam tip: when dividing powers with the same non-zero base, subtract the exponents: \(a^m/a^n=a^{m-n}\).

From (\sqrt{x}=3\sqrt{2}), (x=18), and (x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).

Here \(\left(\frac{3}{5}\right)^{-2}=\frac{25}{9}\) and \(\left(\frac{5}{3}\right)^{-2}=\frac{9}{25}\), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.

The left side is \(3^{2x}\cdot3^{x-1}=3^{3x-1}\), and \(729=3^{6}\). Hence \(3x-1=6\) and \(x=\frac{7}{3}\).

Here (\sqrt{147}=7\sqrt{3}), (2\sqrt{12}=4\sqrt{3}), and (3\sqrt{27}=9\sqrt{3}), so the numerator is (12\sqrt{3}). Therefore, the value should be (12).