Irrational numbers

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.

Step 1: (\sqrt{8}=2\sqrt{2}), (\sqrt{32}=4\sqrt{2}), and (\sqrt{128}=8\sqrt{2}). Step 2: The sum is ((1+2+4+8)\sqrt{2}=15\sqrt{2}). Step 3: Recognize the pattern of perfect-square factors.

Step 1: Assuming (\sqrt{5}=\frac{p}{q}) and squaring gives (p^2=5q^2). Step 2: This makes both (p) and (q) divisible by (5), contradicting that they are coprime. Step 3: Finding a common factor is the key contradiction in the proof.

Step 1: Both numbers are positive, so compare their squares. Step 2: ((4\sqrt{3})^2=48) and ((3\sqrt{5})^2=45), so (4\sqrt{3}) is greater. Step 3: Squaring is safe for comparing positive surds.

Step 1: (x^2=11+7+2\sqrt{77}=18+2\sqrt{77}). Step 2: Therefore (x^2-18=2\sqrt{77}), which is irrational. Step 3: In the square of a sum of different surds, the middle term is the key.

Step 1: (\sqrt{12}=2\sqrt{3}) and (\sqrt{3}) are both irrational. Step 2: Their product is (\sqrt{36}=6), which is rational, and their sum is (3\sqrt{3}), which is irrational. Step 3: Check the nature of the sum and product separately.

Step 1: (x=\sqrt{\frac{3}{2}}), which is irrational because (\frac{3}{2}) is not a perfect square of a rational number. Step 2: (x^2=\frac{3}{2}), which is rational. Step 3: The square of an irrational number can sometimes be rational.

Step 1: The conjugate of the denominator is (\sqrt{5}-\sqrt{2}). Step 2: The denominator becomes (5-2=3), so the form is (\frac{\sqrt{5}-\sqrt{2}}{3}). Step 3: For a sum of two surds, the conjugate changes the sign between them.

Step 1: In the square of three terms, pairwise products appear along with individual squares. Step 2: Thus (a^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}). Step 3: While squaring a sum of many surds, write all pairwise products.

Step 1: (25) and (49) are both perfect squares. Step 2: (\sqrt{25}+\sqrt{49}=5+7=12), which is rational. Step 3: For a rational sum, check both square roots separately.

Step 1: (x^2=5+2\sqrt{6}). Step 2: Using the identity (x^2+\frac{1}{x^2}=10), we get (x^4-10x^2+1=0). Step 3: In such questions, recognize the relation between (x) and its conjugate reciprocal.

Step 1: (\sqrt{a}\times\sqrt{b}=\sqrt{ab}). Step 2: For (a=6,b=15), (ab=90), which is not a perfect square, so (\sqrt{90}) is irrational. Step 3: In multiplication, the key check is whether the product inside the root is a perfect square.

Step 1: ((5-\sqrt{24})(5+\sqrt{24})=25-24=1). Step 2: Therefore (5+\sqrt{24}) is the reciprocal of (5-\sqrt{24}). Step 3: If conjugates multiply to (1), the reciprocal is directly the conjugate.

Step 1: A non-zero rational multiplier does not remove irrationality. Step 2: If the product were rational, the irrational number would become rational, a contradiction. Step 3: Be careful with universal statements about two irrational numbers.

Step 1: (\sqrt{8}=2\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Step 2: (x=5\sqrt{2}), so (\frac{x}{\sqrt{2}}=5). Step 3: Division is easier after combining like surds.

Step 1: All terms are positive and (\sqrt{6}>0). Step 2: Since (\sqrt{12}=2\sqrt{3}) and (\sqrt{6}>\sqrt{3}), the sum (\sqrt{3}+\sqrt{6}) is greater than (2\sqrt{3}). Step 3: For comparison, convert what you can and use positivity.