Irrational numbers

Step 1: (a^2-6a=a(a-6)). Step 2: (a-6=\sqrt{5}-3), so (a(a-6)=(3+\sqrt{5})(\sqrt{5}-3)=5-9=-4). Step 3: Recognize the hidden conjugate form.

Step 1: Reappearance of a digit alone is not recurrence unless a fixed block repeats. Step 2: Here the number of zeros changes, so the decimal is non-recurring. Step 3: Identify a non-terminating non-recurring decimal as irrational.

Step 1: This is a conjugate product. Step 2: ((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=13-12=1). Step 3: In such forms, identify the difference of squares before expanding.

Step 1: (\sqrt{25}=5) and (\sqrt{9}=3). Step 2: The difference is (5-3=2), which is rational. Step 3: A simple way to get a rational difference is to use square roots of perfect squares.

Step 1: (x^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}). Step 2: Subtracting (2\sqrt{6}) leaves (5). Step 3: After squaring, cancel like irrational terms.

Step 1: (\sqrt{75}=5\sqrt{3}) and (\sqrt{12}=2\sqrt{3}). Step 2: The numerator becomes (3\sqrt{3}), so division gives (3). Step 3: Subtract first, then divide by the denominator.

Step 1: (\frac{1}{\sqrt{5}-2}=\sqrt{5}+2), because ((\sqrt{5}-2)(\sqrt{5}+2)=1). Step 2: Hence (x+\frac{1}{x}=(\sqrt{5}-2)+(\sqrt{5}+2)=2\sqrt{5}). Step 3: When conjugates multiply to (1), the reciprocal is immediate.

Step 1: (2=\sqrt{4}) and (3=\sqrt{9}). Step 2: (5) and (8) lie between (4) and (9) and are not perfect squares. Therefore (\sqrt{5}) and (\sqrt{8}) are irrational numbers between (2) and (3). Step 3: Non-perfect squares between two square numbers give such pairs.

Step 1: (a-b=2\sqrt{2}) and (a+b=2\sqrt{3}). Step 2: (\frac{a-b}{a+b}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}). Step 3: Do not forget to rationalize the denominator at the end.

Step 1: For (a=7,b=2), (\sqrt{7}+\sqrt{2}) is irrational. Step 2: The product is ((\sqrt{7})^2-(\sqrt{2})^2=7-2=5), which is rational. Step 3: A conjugate product can give a rational result even when the sum is irrational.

Step 1: (x-\sqrt{2}=\sqrt{5}). Step 2: Therefore ((x-\sqrt{2})^2=(\sqrt{5})^2=5). Step 3: Simplify inside the bracket before expanding the square.

Step 1: ((\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})=3-2=1). Step 2: Therefore (\sqrt{3}-\sqrt{2}) is its reciprocal. Step 3: In reciprocals, keep the order and sign of the conjugate carefully.

Step 1: (x-1=\sqrt{2}+\sqrt{3}). Step 2: (\sqrt{2}+\sqrt{3}) is irrational, because assuming it rational and squaring would force (\sqrt{6}) to be rational. Step 3: Check sums of different surds carefully.

Step 1: (\sqrt{96}=4\sqrt{6}), (\sqrt{54}=3\sqrt{6}), and (\sqrt{24}=2\sqrt{6}). Step 2: (4\sqrt{6}-3\sqrt{6}+2\sqrt{6}=3\sqrt{6}), so the correct value is (3\sqrt{6}). Step 3: Match the options with your simplified result carefully.

Step 1: (y-x=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{5})). Step 2: (\sqrt{5}) cancels and (\sqrt{7}-\sqrt{3}) remains, which is irrational. Step 3: After like terms cancel, check the nature of the remaining surds.

Step 1: (3+\sqrt{2}) and (3-\sqrt{2}) both contain an irrational part, so both are irrational. Step 2: Their product is (9-2=7), which is rational. Step 3: Conjugate irrational numbers can have a rational product.

Step 1: For (x=\sqrt{2}), (x^2=2). Step 2: The denominator (x^2-2=0), so the fraction is undefined. Step 3: Before evaluating a fraction, always check whether the denominator becomes zero.

Step 1: (\sqrt{45}=3\sqrt{5}). Step 2: The difference is (3\sqrt{5}-2\sqrt{5}=\sqrt{5}), which is irrational. Step 3: For like surds, subtract only the coefficients.

Step 1: (x^2=5+2+2\sqrt{10}=7+2\sqrt{10}). Step 2: Thus (x^2-7=2\sqrt{10}), and its square is (40). Step 3: First isolate the irrational part, then square it.

Step 1: (11) is a prime number. Step 2: A prime number is not a perfect square, so (\sqrt{11}) is irrational. Step 3: For the square root of a prime, use the non-perfect-square idea directly.

Step 1: (x) and (y) are conjugates. Step 2: (x^2+y^2=2(4^2+15)=2(31)=62). Step 3: In the sum of squares of conjugates, irrational terms cancel.

Step 1: (\sqrt{2}), (\sqrt{3}), and (\sqrt{6}) are linked to different non-perfect squares. Step 2: Their irrational parts do not cancel through ordinary addition, so the sum is not rational. Step 3: Avoid false identities such as (\sqrt{a+b}=\sqrt{a}+\sqrt{b}).

Step 1: ((x-2)=\sqrt{7}) and ((x+2)=\sqrt{7}+4). Step 2: The product is (\sqrt{7}(\sqrt{7}+4)=7+4\sqrt{7}). Step 3: Before applying an identity directly, substitute the given value of (x) carefully.

Step 1: (\sqrt{18}=3\sqrt{2}), (\sqrt{50}=5\sqrt{2}), and (\sqrt{98}=7\sqrt{2}). Step 2: (1\sqrt{2}+3\sqrt{2}-5\sqrt{2}+7\sqrt{2}=6\sqrt{2}). Step 3: In long surd expressions, write the coefficients separately and add them.

Step 1: ((\sqrt{6}+\sqrt{5})(\sqrt{6}-\sqrt{5})=1), so (\frac{1}{x}=\sqrt{6}-\sqrt{5}). Step 2: (x+\frac{1}{x}=2\sqrt{6}), hence (x^3+\frac{1}{x^3}=(2\sqrt{6})^3-3(2\sqrt{6})=42\sqrt{6}). Step 3: In cube-type questions, finding (x+\frac{1}{x}) first is the easier method.