Irrational numbers

Step 1: ((u+v)^2-(u-v)^2=4uv). Step 2: Here (u=\sqrt{7}) and (v=\sqrt{2}), so the value is (4\sqrt{14}). Step 3: Using the identity makes the expansion shorter.

Step 1: Rationalize the denominator of (\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}). Step 2: The numerator becomes ((\sqrt{6}+\sqrt{2})^2=8+4\sqrt{3}), and the denominator is (6-2=4), so the value is (2+\sqrt{3}). Step 3: Multiplying by the conjugate is effective in such quotients.

Step 1: (\sqrt{11}) is about (3.316), and (2\sqrt{3}) is about (3.464). Step 2: (3.4) lies between them. Step 3: For close values, estimating to two decimal places is helpful.

Step 1: (x^2=2). Step 2: Therefore (x^4=(x^2)^2=4), and the value is (4-8+4=0). Step 3: For powers of a surd, first find (x^2).

Step 1: From (p^2=2q^2), first (p^2) is even and hence (p) is even. Step 2: After writing (p=2k), we get (q^2=2k^2), so (q) is even. Step 3: Skipping this order makes the proof incomplete.

Step 1: (a^2-2a=a(a-2)). Step 2: (a-2=\sqrt{5}-1), so (a(a-2)=(1+\sqrt{5})(\sqrt{5}-1)=4). Step 3: Recognizing the hidden conjugate form is a quick method.

Step 1: (\sqrt{48}=4\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Step 2: (4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}). Step 3: For like surds, work with the coefficients.

Step 1: (x-\sqrt{3}=\sqrt{2}) and (x-\sqrt{2}=\sqrt{3}). Step 2: Their product is (\sqrt{2}\times\sqrt{3}=\sqrt{6}). Step 3: Simplify the small brackets first.

Step 1: (\sqrt{8}=2\sqrt{2}), (\sqrt{18}=3\sqrt{2}), and (\sqrt{32}=4\sqrt{2}). Step 2: The total is (1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}=10\sqrt{2}). Step 3: In ordered surds, identify the coefficient pattern.

Step 1: First observe the common structure and take (a=\sqrt{7}+\sqrt{5}) and (b=\sqrt{7}-\sqrt{5}). Step 2: (\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}). Here (a^2+b^2=24) and (ab=2), so (x=12). Step 3: For fractions with conjugate surds, use substitution instead of expanding everything directly.

Step 1: Multiply by (\sqrt{10}+\sqrt{6}) to rationalize the denominator. Step 2: The numerator becomes ((\sqrt{10}+\sqrt{6})^2=16+2\sqrt{60}) and the denominator is (10-6=4), so the value is (4+\sqrt{15}). Step 3: In conjugate fractions, clear the denominator first.

Step 1: (x) and (y) are conjugates. Step 2: In ((\sqrt{5}+\sqrt{2})^2+(\sqrt{5}-\sqrt{2})^2), the middle irrational terms cancel and the value is (2(5+2)=14). Step 3: When adding squares of conjugates, the middle terms vanish.

Step 1: Write (\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}). Step 2: The numerator is (5\sqrt{3}), so (\frac{5\sqrt{3}}{\sqrt{3}}=5). Step 3: Combine like surds before division.

Step 1: (\sqrt{9}=3) and (\sqrt{16}=4). Step 2: Their sum is (3+4=7), so this pair satisfies the condition. Step 3: To get an integer sum, first check the perfect-square options.

Step 1: (\frac{1}{2+\sqrt{3}}=2-\sqrt{3}). Step 2: Hence (x+\frac{1}{x}=4), so (x^2+\frac{1}{x^2}=4^2-2=14). Step 3: Finding (x+\frac{1}{x}) first saves long calculation.

Step 1: This decimal does not terminate. Step 2: The number of zeros keeps changing, so no fixed repeating block is formed. Step 3: A non-terminating non-recurring decimal is irrational.

Step 1: Use (a^2-b^2=(a-b)(a+b)). Step 2: (a-b=2\sqrt{2}) and (a+b=2\sqrt{6}), so the product is (4\sqrt{12}=8\sqrt{3}). Step 3: Identities make the solution quicker and cleaner.

Step 1: (3+\sqrt{8}=3+2\sqrt{2}) is irrational. Step 2: Its reciprocal is (3-\sqrt{8}), because ((3+\sqrt{8})(3-\sqrt{8})=1). Hence the sum is (6), which is rational. Step 3: When conjugates multiply to (1), the reciprocal is easy to identify.

Step 1: (\sqrt{18}=3\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{8}=2\sqrt{2}). Step 2: (3\sqrt{2}+5\sqrt{2}-2\sqrt{2}=6\sqrt{2}). Step 3: Keep the signs carefully while adding or subtracting coefficients.

Step 1: Use ((a-b)^2=a^2-2ab+b^2). Step 2: (x^2=7-2\sqrt{21}+3=10-2\sqrt{21}). Step 3: Do not forget the negative sign of the middle term in the square of a difference.