Irrational numbers

Step 1: (2) is rational. So (2+\sqrt{n}) is irrational when (\sqrt{n}) is irrational. Step 2: (40) is not a perfect square, so (\sqrt{40}) is irrational. Step 3: First eliminate perfect squares from the given integers.

Step 1: (\sqrt{75}=5\sqrt{3}) and (\sqrt{27}=3\sqrt{3}). Step 2: The sum is (5\sqrt{3}+3\sqrt{3}=8\sqrt{3}). Step 3: Do not combine separate square roots directly into one root.

Step 1: ((\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}). Step 2: This is (7+2\sqrt{10}), which has an irrational part. Step 3: When squaring a sum of two different surds, pay attention to the middle term.

Step 1: (4=\sqrt{16}) and (5=\sqrt{25}). Step 2: (17), (20), and (24) lie between (16) and (25), but (26) is greater than (25). Step 3: For positive square roots, comparing squares is easier.

Step 1: (3x-2=3\sqrt{7}-2). Step 2: (3\sqrt{7}) is irrational, and subtracting a rational number keeps it irrational. Step 3: A non-zero rational multiple of a surd remains irrational.

Step 1: (\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}). Step 2: This is (\sqrt{9}=3), which is rational. Step 3: In division, simplifying the radicals together is a quick method.

Step 1: (\sqrt{8}=2\sqrt{2}). Step 2: Therefore (\sqrt{8}-2\sqrt{2}=0), which is rational. Step 3: Sometimes terms that look irrational cancel completely.

Step 1: (x^2-2x=x(x-2)). Step 2: With (x=1+\sqrt{2}), (x-2=\sqrt{2}-1), so the product ((1+\sqrt{2})(\sqrt{2}-1)=1). Step 3: Recognizing conjugate-like forms makes calculation shorter.

Step 1: (\sqrt{3}) is about (1.732), and (2\sqrt{3}) is about (3.464). Step 2: (2) lies between these two values. Step 3: For comparison, you may use estimation or squaring.

Step 1: Simplify each option first. Step 2: (\sqrt{6}+\sqrt{24}=\sqrt{6}+2\sqrt{6}=3\sqrt{6}), which is irrational. Step 3: Radicals may cancel in multiplication, but not always in addition.

Step 1: Since (m) is not a perfect square, (\sqrt{m}) is irrational. Step 2: (4) is rational, and adding it to an irrational number gives an irrational number. Step 3: Connect the non-perfect-square condition directly with the nature of the square root.

Step 1: (\sqrt{a}\times\sqrt{b}=\sqrt{ab}). Step 2: For (a=3,b=12), (ab=36), so (\sqrt{36}=6), which is rational. Step 3: Check whether the product inside the radical becomes a perfect square.

Step 1: The conjugate of the denominator is (\sqrt{2}-1). Step 2: (\frac{2}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2(\sqrt{2}-1)}{2-1}=2\sqrt{2}-2). Step 3: Choosing the correct conjugate sign is very important.

Step 1: Use ((a-b)^2=a^2-2ab+b^2). Step 2: (x^2=5-2\sqrt{10}+2=7-2\sqrt{10}). Step 3: Do not forget the negative sign in the middle term when squaring a difference.

Step 1: (\sqrt{8}=2\sqrt{2}), so (x) and (y=-2\sqrt{2}) are both irrational. Step 2: Their sum is (2\sqrt{2}-2\sqrt{2}=0), which is rational. Step 3: Opposite irrational terms can give a rational sum.

Step 1: (\sqrt{2}) is irrational, and dividing by non-zero rational (2) keeps it irrational. Step 2: (\frac{\sqrt{2}}{2}) is about (0.707), so it lies between (0) and (1). Step 3: Check both the interval condition and the nature of the number.

Step 1: (\sqrt{12}=2\sqrt{3}) and (\sqrt{27}=3\sqrt{3}). Step 2: The total sum is (\sqrt{3}+2\sqrt{3}+3\sqrt{3}=6\sqrt{3}). Step 3: Converting all terms into like surds makes addition easy.

Step 1: (36) is a perfect square. Step 2: (\sqrt{36}=6), which is rational, so the statement for (p=36) is incorrect. Step 3: Checking perfect squares is the safest way to decide the nature of a square root.

Step 1: View (ab) as ((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})). Step 2: This equals ((\sqrt{3})^2-(\sqrt{2})^2=3-2=1). Step 3: Since addition order does not change the sum, recognize the conjugate form.

Step 1: Use ((a+b)^2-(a-b)^2=4ab). Step 2: Here (a=3) and (b=\sqrt{2}), so (A=4\times3\times\sqrt{2}=12\sqrt{2}), which is irrational. Step 3: In such questions, use the identity instead of expanding both squares fully.