Operations on real numbers and the laws of exponents

Here \(\frac{1}{a}=2-\sqrt{3}\), so \(a+\frac{1}{a}=4\) and \(a^{2}+\frac{1}{a^{2}}=4^{2}-2=14\). In exams, use the identity \(\left(a+\frac{1}{a}\right)^{2}\).

The first product is \(7-5=2\), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.

Since (125=5^{3}), (x=3), and since (32=2^{5}), (y=5), so (x+y=8). In exams, write numbers as powers of their prime bases.

Here (3^{-2}+3^{-1}=\frac{1}{9}+\frac{1}{3}=\frac{4}{9}), and (3^{-3}=\frac{1}{27}), so the value is (12). In exams, convert negative powers into fractions.

We use \(x^{2}-4=x^{2}-2^{2}=(x-2)(x+2)\). In exams, remember the difference of squares form \((a^{2}-b^{2})\).

((\sqrt{6}+\sqrt{2})^{2}=6+2+2\sqrt{12}=8+4\sqrt{3}). In exams, do not miss the middle term of ((a+b)^{2}).

For the same base (10), the exponent is (2-5+4=1), so (r=10^{1}=10). In exams, add exponents during multiplication.

Inside, \(a^{3-(-1)}b^{-2-2}=a^{4}b^{-4}\), and squaring gives \(a^{8}b^{-8}\). In exams, watch the sign when subtracting negative exponents.

(\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\frac{1}{2}), so (u^{-2}=4). In exams, simplify the radical first.

The zero-exponent law states that \(a^{0}=1\) for every nonzero number, so option A is correct. Option B is false because \(a^{0}\) equals 1, not 0. In option C, \(a^{-1}=\frac{1}{a}\), which is not generally equal to \(a\); and in option D, \(a^{1}=a\). Exam tip: Apply \(a^{0}=1\) only when \(a\ne0\); do not include \(0^{0}\) in this rule.

Since (\frac{1}{2-\sqrt{3}}=2+\sqrt{3}), (\frac{1}{x}+x=(2+\sqrt{3})+(2-\sqrt{3})=4). In exams, identify conjugate numbers quickly.

Since (6^{5}=2^{5}\cdot3^{5}), (\frac{2^{5}3^{5}}{2^{3}3^{4}}=2^{2}\cdot3). In exams, split a composite base into prime bases.

The left side has exponents (2n) and (3n), so \(2n=8\) and \(3n=12\), giving \(n=4\). In exams, match exponents of both variables.

Since (\sqrt[3]{64}=4) and (\sqrt[3]{x^{6}}=x^{2}), the answer is (4x^{2}). In exams, divide the exponent by (3) for cube roots.

\((\sqrt{11}-\sqrt{2})^{2}=11+2-2\sqrt{22}=13-2\sqrt{22}\). In exams, include both \(+b^{2}\) and (-2ab) in \((a-b)^{2}\).

From (2^{a}=2^{3}), (a=3), and from (3^{b}=3^{4}), (b=4), so (a^{b}=3^{4}=81). In exams, compare powers using equal bases.

(\frac{x^{5}-x^{3}}{x^{3}}=\frac{x^{3}(x^{2}-1)}{x^{3}}=x^{2}-1). In exams, take out the common factor first.

Here \(16^{-\frac{3}{4}}=(2^{4})^{-\frac{3}{4}}=2^{-3}\) and \(8^{\frac{2}{3}}=(2^{3})^{\frac{2}{3}}=2^{2}\), so the value is \(2^{-1}=\frac{1}{2}\). In exams, multiply powers of powers.

Since ((2+\sqrt{3})^{2}=4+3+4\sqrt{3}=7+4\sqrt{3}), (\sqrt{A}=2+\sqrt{3}). In exams, recognize the form ((a+b)^{2}).

Here (x^{-1}+y^{-1}=\frac{x+y}{xy}) and ((xy)^{-1}=\frac{1}{xy}), so division gives (x+y). In exams, converting negative exponents to fractions is safer.