Real Numbers

Since \(125=25\times5\), \(\sqrt{125}=\sqrt{25\times5}=5\sqrt{5}\). Therefore, \(\sqrt{125}-\sqrt{5}=5\sqrt{5}-\sqrt{5}=4\sqrt{5}\), so option B is correct. \(\sqrt{120}\) is not equal to \(\sqrt{125}-\sqrt{5}\), because subtraction of square roots cannot be combined into one square root. Exam tip: simplify each surd by identifying its largest perfect-square factor first.

Given \(x=\sqrt{14}\), we get \(x^2=(\sqrt{14})^2=14\). Therefore, \(x^2-10=14-10=4\), so option D is correct. The value 14 is only \(x^2\); 10 still has to be subtracted. Exam tip: squaring a square root gives the original non-negative number.

Since \(18=9\times2\) and \(8=4\times2\), \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\). Therefore, \(\sqrt{18}+\sqrt{8}=3\sqrt{2}+2\sqrt{2}=5\sqrt{2}\). \(3\sqrt{2}\) is only the simplified form of \(\sqrt{18}\), not of the sum. Exam tip: simplify each surd by taking out perfect-square factors before adding like surds.

\(\sqrt{81}=9\) and \(\sqrt{9}=3\). Therefore, \(\frac{\sqrt{81}}{\sqrt{9}}=\frac{9}{3}=3\). Hence, option B is correct. The value 9 is only the square root of the numerator; division by the denominator is still required. Exam tip: evaluate each square root first, then simplify the fraction.

The set of real numbers contains all rational numbers and all irrational numbers. A rational number can be written in the form \(p/q\), where \(q\ne0\), whereas an irrational number such as \(\sqrt{2}\) cannot be written in this form. Integers and natural numbers are only subsets of rational numbers, so they do not give the complete classification of real numbers. Exam tip: remember that real numbers = rational numbers + irrational numbers.

(\sqrt{20}=2\sqrt{5}) and (\sqrt{45}=3\sqrt{5}), so the sum is (5\sqrt{5}). Simplify surds before adding.

In simplest form the denominator contains 3 along with 2, so the decimal will not terminate. Check prime factors of the denominator.

Since \(27=9\times3\) and \(12=4\times3\), \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Therefore, \(x=3\sqrt{3}-2\sqrt{3}=\sqrt{3}\). \(2\sqrt{3}\) is only the simplified value of \(\sqrt{12}\), not the value of the difference. Exam tip: simplify both surds to like radicals before subtracting.

Here, \(x=\sqrt{2}+\sqrt{3}\). Therefore, \(x^2=(\sqrt{2}+\sqrt{3})^2=2+3+2\sqrt{2}\sqrt{3}=5+2\sqrt{6}\). Hence, option A is correct. In option B, the coefficient 2 in the middle term \(2ab\) has been omitted. Exam tip: When squaring two surds, always include the middle term \(2ab\).

Since \(72=36\times2\), and \(36\) is a perfect square, \(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\). Therefore, \(6\sqrt{2}\) is correct. For example, \(3\sqrt{2}\) squares to \(18\), not \(72\). Exam tip: to simplify a surd, first identify the greatest perfect-square factor of the number.

Given \(a=\sqrt{5}-2\), adding 2 gives \(a+2=\sqrt{5}\). Therefore, \((a+2)^2=(\sqrt{5})^2=5\). Option 10 would result from incorrectly evaluating the square of \(\sqrt{5}\). Exam tip: use the identity \((\sqrt{x})^2=x\).

Since \(18=9\times2\), \(\sqrt{18}=3\sqrt{2}\); and since \(50=25\times2\), \(\sqrt{50}=5\sqrt{2}\). Therefore, \(\sqrt{18}+\sqrt{50}=3\sqrt{2}+5\sqrt{2}=8\sqrt{2}\). The option \(7\sqrt{2}\) would result from adding the coefficients incorrectly. Exam tip: first take out perfect-square factors from surds, then combine like surds.

\(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\). Therefore, \(a+b=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\). \(2\sqrt{3}\) is only the simplified value of \(b\), not the sum. Exam tip: simplify surds to like radicals before adding them.

Given \(\sqrt{x}=9\), squaring both sides gives \(x=9^2=81\). Hence, 81 is correct. The number 9 is the value of \(\sqrt{x}\), not of \(x\). Exam tip: After squaring a square-root equation, check the value in the original equation.

Since \(98=49\times2\) and \(8=4\times2\), \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\). Therefore, \(\sqrt{98}-\sqrt{8}=7\sqrt{2}-2\sqrt{2}=5\sqrt{2}\), so option C is correct. \(4\sqrt{2}\) would result only if the coefficients differed by 4. Exam tip: first take perfect-square factors out of each surd, then subtract like surd terms.

Given \(x=\sqrt{2}-1\), we get \(x(\sqrt{2}+1)=(\sqrt{2}-1)(\sqrt{2}+1)\). This matches \((a-b)(a+b)=a^2-b^2\), so its value is \((\sqrt{2})^2-1^2=2-1=1\). Option 2 would result from forgetting to subtract \(1^2\). Exam tip: identify conjugate factors and use the difference-of-squares identity.

Since \(45=9\times5\) and \(80=16\times5\), \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\). Therefore, \(\sqrt{45}+\sqrt{80}=3\sqrt{5}+4\sqrt{5}=7\sqrt{5}\). A result such as \(9\sqrt{5}\) can arise from simplifying the square roots incorrectly. Exam tip: first factor out perfect squares, such as \(9\) or \(16\), from under each radical.

\(\sqrt{2}\approx 1.414\), \(\frac{3}{2}=1.5\), and \(\sqrt{3}\approx 1.732\). All three are greater than 1, so 1 is the smallest number. \(\sqrt{2}\) may seem like the closest distractor, but it is still greater than 1. Exam tip: use approximate decimal values to compare square roots quickly.

Given \(a=\sqrt{8}\), we get \(a^2=(\sqrt{8})^2=8\). Squaring the principal square root of a positive number gives back the number itself. Option 4 is not the square of \(\sqrt{8}\); it is the value of \(\sqrt{16}\). Exam tip: use \((\sqrt{x})^2=x\) and write the radicand directly.

Since \(200=100\times2\) and \(100\) is a perfect square, \(\sqrt{200}=\sqrt{100}\times\sqrt{2}=10\sqrt{2}\). The option \(5\sqrt{2}\) squares to \(50\), so it is not equal to \(\sqrt{200}\). Exam tip: To simplify a surd, first identify the greatest perfect-square factor of the number.