Irrational numbers

(\sqrt{75}=5\sqrt{3}) and (\sqrt{192}=8\sqrt{3}), so (s=13\sqrt{3}). Dividing by (\sqrt{3}) gives (13).

Multiplying by the conjugate gives (\frac{(\sqrt{7}-\sqrt{3})^2}{7-3}). So the answer is (\frac{5-\sqrt{21}}{2}).

Option C is correct. If \(r+x\) were rational, then \(x=(r+x)-r\) would also be rational, a contradiction. For A, \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims using a counterexample.

Since 2, 3 and 5 are not perfect squares, \(\sqrt2\), \(\sqrt3\) and \(\sqrt5\) are all irrational. The other options contain \(\sqrt{49}=7\), \(\sqrt2\times\sqrt8=\sqrt{16}=4\), or \(\frac{\sqrt{12}}{\sqrt3}=\sqrt4=2\), which are rational. Exam tip: a square root of a perfect square is rational.

Area is ((5+\sqrt{11})(5-\sqrt{11})=25-11=14). Multiplying conjugate dimensions can give a rational area.

(\sqrt{500}=10\sqrt{5}), (\sqrt{320}=8\sqrt{5}), and (\sqrt{180}=6\sqrt{5}). So the result is (10\sqrt{5}-8\sqrt{5}+6\sqrt{5}=8\sqrt{5}).

The first term becomes (\frac{\sqrt{8}-\sqrt{6}}{2}) and the second becomes (\frac{\sqrt{8}+\sqrt{6}}{2}). Their sum is (\sqrt{8}).

Both \(\sqrt{5}\) and \(3-\sqrt{5}\) are irrational, but their sum is \(\sqrt{5}+3-\sqrt{5}=3\), which is rational. Thus the claim is false. Exam tip: check whether irrational terms cancel.

Both \(\sqrt{7}\) and \(-\sqrt{7}\) are irrational, but they are additive inverses, so their sum is \(0\), a rational number. Thus the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.

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.

The governing concept is the difference-of-squares identity for conjugate expressions: (a+b)(a−b)=a²−b². Let a=√45 and b=√20. The required value is therefore (√45)²−(√20)²=45−20=25, making option A correct. Principal square roots are non-negative, and squaring each one returns its radicand exactly. Option B would result from reversing the subtraction order, but the given factors are arranged as a+b followed by a−b. Option C incorrectly adds the radicands. Option D has no valid connection with the identity and comes from an unrelated manipulation. One may also simplify the radicals to 3√5 and 2√5; the factors become 5√5 and √5, whose product is again 25.

\(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\), 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}).