Irrational numbers

If \(r+x\) were rational, subtracting the rational number \(r\) would make \(x\) rational, a contradiction. Hence A is correct. Option B is also false because the difference remains irrational. Use contradiction in such questions.

Since (14<16<18), (\sqrt{16}) lies between them. For positive square roots compare the numbers inside.

If \(a+bx\) were rational, then \(bx\) would be rational; since \(b\ne0\), \(x=\frac{bx}{b}\) would be rational, a contradiction. But \(x^2\) can be rational, for example when \(x=\sqrt2\). Exam tip: adding a rational number to an irrational number keeps it irrational.

(\sqrt{200}=10\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the numerator is (4\sqrt{2}). Dividing gives (4).

The square of a square root gives the number inside. So \(\left(\sqrt{11+\sqrt{30}}\right)^2=11+\sqrt{30}\).

Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence, the word “always” makes the statement false. Exam tip: one counterexample is enough to disprove an “always” statement.

\(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\), which is rational. Hence, a quotient of two irrational numbers need not be irrational. Option B simplifies to \(\sqrt{3}\), which is irrational. Exam tip: combine the radicals first, then simplify the number inside the square root.

Multiplying by the conjugate gives numerator (13+2\sqrt{22}) and denominator (11-2=9). So the simplified form is (\frac{13+2\sqrt{22}}{9}).

If \(a+r\) were rational, then \(a=(a+r)-r\) would also be rational, contradicting that \(a\) is irrational. Hence the sum is irrational. Exam tip: adding or subtracting a rational number preserves irrationality.

(\sqrt{605}=11\sqrt{5}), (\sqrt{245}=7\sqrt{5}), and (\sqrt{125}=5\sqrt{5}). Therefore the result is (9\sqrt{5}).

The direct answer is A, 192. Simplify the radicals: √27 = √(9×3) = 3√3 and √75 = √(25×3) = 5√3. Their sum is 8√3. Squaring gives (8√3)^2 = 64×3 = 192, so option A is correct. Option B, 128, uses the wrong final multiplication and does not equal the expression. Option C, 16√3, is not the square of 8√3; squaring removes the remaining radical and gives a rational number. Option D, 300, results from an incorrect expansion. As a second check, (√27+√75)^2 = 27+75+2√(27×75) = 102+2√2025 = 102+90 = 192. Memory cue: when both radicals contain the same remaining factor, add their coefficients first, then square.

Multiplying by the conjugate makes the denominator (14-5=9). So (\frac{3(\sqrt{14}-\sqrt{5})}{9}=\frac{\sqrt{14}-\sqrt{5}}{3}).

(\sqrt{18}=3\sqrt{2}) and (\sqrt{72}=6\sqrt{2}) so the numerator is (9\sqrt{2}). Dividing by (\sqrt{2}) gives (9).

(\sqrt{12}=2\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Therefore (a-b=3\sqrt{3}).

Since \(12=4\times3\), \(\sqrt{12}=2\sqrt{3}\). Thus the difference is \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\), which is irrational. Exam tip: never use \(\sqrt a-\sqrt b=\sqrt{a-b}\).

Multiplying by the conjugate makes the denominator (13-5=8). So (\frac{4(\sqrt{13}+\sqrt{5})}{8}=\frac{\sqrt{13}+\sqrt{5}}{2}).

Area is ((\sqrt{11}+\sqrt{6})(\sqrt{11}-\sqrt{6})=11-6=5). Conjugate dimensions can give a rational area.

(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6), which is rational. Check whether multiplication creates a perfect square.