Irrational numbers

Multiplying by the conjugate gives denominator (16-7=9) and numerator ((4+\sqrt{7})^2=23+8\sqrt{7}). So the correct form is (\frac{23+8\sqrt{7}}{9}).

\(0.\overline{3}=1/3\), so it is non-terminating but repeating and hence rational. In contrast, \(\sqrt{2}\) and \(\pi\) are irrational. Exam tip: a non-terminating repeating decimal is always rational.

(\sqrt{147}=7\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Therefore the result is (5\sqrt{3}).

Multiplying by the conjugate (\sqrt{13}-3) makes the denominator (13-9=4). So (p=\frac{\sqrt{13}-3}{4}).

Multiplying the same square root by itself gives the number inside. Therefore the value is (9+\sqrt{20}).

Option C is correct. For \(a=\sqrt{2}, b=-\sqrt{2}\), the sum is 0 and \(ab=-2\) is rational. But with \(b=1-\sqrt{2}\), the sum is 1 while \(ab=\sqrt{2}-2\) is irrational. Exam tip: verify a claim using contrasting examples.

The conjugate of the denominator is (\sqrt{5}+2) and the denominator becomes (1). So the value is ((\sqrt{5}+2)^2=9+4\sqrt{5}).

Multiplying by the denominator conjugate gives denominator (8-3=5). The numerator is ((\sqrt{8}+\sqrt{3})^2=11+4\sqrt{6}).

Assume \(a+b\) is rational. Then \(a=(a+b)-b\) would be the difference of two rational numbers and hence rational, a contradiction. Therefore \(a+b\) is irrational. Exam tip: rational ± irrational is always irrational.

Since (10+\sqrt{21}) is positive, its square root is real. Squaring it gives the inside number.

If \(x+r\) were rational, then \(x=(x+r)-r\) would be the difference of two rational numbers and hence rational, a contradiction. Thus A is correct. Exam tip: rational ± irrational is always irrational.

Multiplying by the conjugate makes the denominator (15-6=9). So (\frac{6(\sqrt{15}+\sqrt{6})}{9}=\frac{2(\sqrt{15}+\sqrt{6})}{3}).

(\sqrt{50}=5\sqrt{2}) and (\sqrt{98}=7\sqrt{2}), so the sum is (12\sqrt{2}). Its square is (288).

If one irrational number is x and the other is its additive inverse −x, then \(x+(-x)=0\). Since 0 is rational, the sum is rational. Exam tip: for “guaranteed” statements, check whether every allowed case works, not just one example.

\(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so the bracket is \(2\sqrt{5}\). Multiplying by \(\sqrt{5}\) gives (10).

Multiplying by the conjugate makes the denominator (12-7=5). So (\frac{5(\sqrt{12}-\sqrt{7})}{5}=\sqrt{12}-\sqrt{7}).

If \(q\ne0\) is rational and \(x\) is irrational, assuming \(qx\) rational gives \(x=(qx)/q\) rational, a contradiction. B fails because \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims with a counterexample.

The governing concept is the square-root identity √x × √x = x for every non-negative real number x. Here, the quantity inside both identical square-root signs is 6 + √5. Since √5 is approximately 2.236, the radicand is positive, so the identity applies directly: √(6 + √5) × √(6 + √5) = (√(6 + √5))² = 6 + √5. Therefore, option A is correct. Option B incorrectly changes the plus sign to a minus sign, option C treats the expression as 6² + (√5)², and option D has no valid connection with the given product.

(\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). Therefore the result is (13\sqrt{2}).

(\frac{4}{\sqrt{13}+3}=\sqrt{13}-3) because the denominator becomes (13-9=4). Therefore the sum is (2\sqrt{13}).