Operations on real numbers and the laws of exponents

Here \(125^{\frac{2}{3}}=(5)^{2}=25\) and \(25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}\). The product is \(\frac{1}{5}\).

When powers with the same base are multiplied, their exponents are added; hence \(a^m\cdot a^n=a^{m+n}\). In division, exponents are subtracted, not multiplied. Exam tip: check whether the bases are identical first.

We have (\sqrt[3]{343}=7), (\sqrt[3]{a^{15}}=a^{5}), and (\sqrt[3]{b^{12}}=b^{4}). In exams, divide exponents by (3) under a cube root.

Since (x^{12}-4096=(x^{6})^{2}-64^{2}=(x^{6}-64)(x^{6}+64)), cancelling the common factor gives (x^{6}+64).

Since (\frac{1}{8-\sqrt{63}}=8+\sqrt{63}), because (64-63=1). Therefore, (\frac{1}{p}-p=2\sqrt{63}).

Here (5^{-2}+5^{-3}=\frac{1}{25}+\frac{1}{125}=\frac{6}{125}), and (5^{-4}=\frac{1}{625}). Division gives (30).

Using the identity \((a-b)^2=a^2+b^2-2ab\), we get \(x^2=(\sqrt{11}-\sqrt{6})^2=11+6-2\sqrt{66}=17-2\sqrt{66}\). Therefore, \(x^2+2\sqrt{66}=17\). Option 5 results from incorrectly subtracting 6 from 11 while squaring. Exam tip: for \((a-b)^2\), the middle term is always \(-2ab\).

Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).

Since (1024=2^{10}), (16^{x}=2^{4x}) gives (x=\frac{5}{2}), and (32^{y}=2^{5y}) gives (y=2). Hence the sum is (\frac{9}{2}).

When powers with the same positive base are multiplied, their exponents are added; hence \(a^p\cdot a^q=a^{p+q}\). In division, exponents are subtracted, not divided: \(a^p/a^q=a^{p-q}\). Exam tip: multiply → add exponents; divide → subtract.

Since (24^{3}=(2^{3}\cdot3)^{3}=2^{9}\cdot3^{3}), division leaves (2^{3}\cdot3=24), so the correct value is not among the options.

Here (\frac{1}{s}=\sqrt{17}-4), so (s-\frac{1}{s}=8) and (s+\frac{1}{s}=2\sqrt{17}). Thus (s^{2}-\frac{1}{s^{2}}=16\sqrt{17}).

We get \(\left(\frac{125x^{-9}}{64y^{12}}\right)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}\). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).

We use \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\). Thus \(60=6\left(x+\frac{1}{x}\right)\), so the value is (10).

Combining like terms gives \(6b^{-3}+9b^{-3}=15b^{-3}\). Therefore, \(\frac{15b^{-3}}{3b^{-5}}=5b^{-3-(-5)}=5b^2\), so option A is correct. Option B results from subtracting the exponents in the wrong order. Exam tip: when dividing powers with the same non-zero base, subtract the denominator exponent from the numerator exponent: \(b^m/b^n=b^{m-n}\).

From (\sqrt{x}=5\sqrt{2}), (x=50), and (x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).

Here \(\left(\frac{5}{8}\right)^{-2}=\frac{64}{25}\) and \(\left(\frac{8}{5}\right)^{-2}=\frac{25}{64}\). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).

Using the laws of exponents, \\( (7^x)^2=7^{2x}\\), so the left-hand side becomes \\(7^{2x}\cdot7^{x-1}=7^{3x-1}\\). Also, \\(16807=7^5\\). Therefore, \\(3x-1=5\\), giving \\(3x=6\\) and hence \\(x=2\\). Exam tip: when multiplying powers with the same base, add their exponents.

Here (\sqrt{363}=11\sqrt{3}), (2\sqrt{147}=14\sqrt{3}), and (3\sqrt{75}=15\sqrt{3}). The numerator is (12\sqrt{3}), so the value should be (12).

Multiplying both sides by (\sqrt{m}+\sqrt{n}) gives (1=m-n). In exams, apply the conjugate product directly.