Real Numbers

\(288=144\times2=12^2\times2\). Therefore, \(\sqrt{288}=\sqrt{12^2\times2}=12\sqrt{2}\), so option C is correct. In option A, the coefficient is doubled, while options B and D are not equal to \(\sqrt{288}\). Exam tip: factor out the largest perfect-square factor before simplifying a square root.

\(\sqrt{45}=\sqrt{9\times5}=3\sqrt{5}\), so \(2\sqrt{45}=6\sqrt{5}\). Similarly, \(\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}\), so \(3\sqrt{20}=6\sqrt{5}\). Adding the like surd terms gives \(6\sqrt{5}+6\sqrt{5}=12\sqrt{5}\), so option A is correct. An answer such as option B results from adding the coefficients incorrectly. Exam tip: simplify each surd first, then add only terms with the same surd part.

\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\). Therefore, \(4\sqrt{27}-2\sqrt{75}=4(3\sqrt{3})-2(5\sqrt{3})=12\sqrt{3}-10\sqrt{3}=2\sqrt{3}\). Hence, option A is correct. Exam tip: Simplify each surd first, then combine only like surd terms.

By distributive law, ( \sqrt{5}\times3+\sqrt{5}\times\sqrt{5}=3\sqrt{5}+5 ). Be careful in surd multiplication.

Apply the identity \((a+b)^2=a^2+2ab+b^2\). Here, \(a=4\) and \(b=\sqrt{3}\), so \((4+\sqrt{3})^2=4^2+2(4)(\sqrt{3})+(\sqrt{3})^2=16+8\sqrt{3}+3=19+8\sqrt{3}\). Therefore, option A is correct. Remember that \((\sqrt{3})^2=3\), not \(\sqrt{4}\).

( (6-\sqrt{8})^2=36-12\sqrt{8}+8 ). Using ( \sqrt{8}=2\sqrt{2} ), it becomes (44-24\sqrt{2}).

This is a product of conjugate expressions. Using \((a+b)(a-b)=a^2-b^2\), we get \((5+\sqrt{17})(5-\sqrt{17})=5^2-(\sqrt{17})^2=25-17=8\). Therefore, the correct answer is 8. The value 42 results from adding 25 and 17, but this product requires their difference. Exam tip: Whenever you see \((a+b)(a-b)\), apply \(a^2-b^2\) directly.

Since \\(50=25\times2\\), we get \\(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\\). Because \\(\sqrt{2}\\) is irrational, multiplying it by the non-zero rational number 5 still gives an irrational number. Therefore, option A is correct. Option C has the correct simplification but incorrectly classifies the result as rational. Exam tip: take perfect-square factors outside the square root and check whether a non-perfect-square factor remains inside.

Multiplying numerator and denominator by ( \sqrt{2} ) gives ( \frac{5\sqrt{2}}{6} ). The radical should be removed from the denominator.

The governing concept is rationalisation of a denominator containing a surd. First, √9=3, so the expression becomes 1/(√10+3). Multiply numerator and denominator by the conjugate √10−3. The result is [1(√10−3)]/[(√10+3)(√10−3)]. Using the difference-of-squares identity, the denominator becomes (√10)²−3²=10−9=1. Therefore the fraction simplifies to √10−3, so option A is correct. Option B is only the original denominator, not the rationalised value. Option C introduces the incorrect denominator 19. Option D is an equivalent reciprocal-style expression before rationalisation, but its denominator still contains a surd and it is not the required simplified rationalised form. The calculation also confirms the sign and denominator exactly.

To rationalise the denominator, multiply the numerator and denominator by its conjugate, \(\sqrt{11}+\sqrt{2}\). The denominator becomes \((\sqrt{11}-\sqrt{2})(\sqrt{11}+\sqrt{2})=11-2=9\), while the numerator becomes \(\sqrt{11}+\sqrt{2}\). Hence, the rationalised form is \(\frac{\sqrt{11}+\sqrt{2}}{9}\). Exam tip: use the identity \((a-b)(a+b)=a^2-b^2\).

( \sqrt{98}=7\sqrt{2} ), ( \sqrt{162}=9\sqrt{2} ), and ( \sqrt{50}=5\sqrt{2} ). So the result is (11\sqrt{2}).

Since \(44=4\times11\) and \(99=9\times11\), we have \(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\). Therefore, \(2\sqrt{44}+3\sqrt{99}=2(2\sqrt{11})+3(3\sqrt{11})=4\sqrt{11}+9\sqrt{11}=13\sqrt{11}\). Hence, option B is correct. Options A and C use incorrect coefficients, while option D changes the radicand incorrectly. Exam tip: simplify each radical first, then combine like radicals.

Since \(112=16\times7\), \(\sqrt{112}=4\sqrt{7}\). Also, \(175=25\times7\), so \(2\sqrt{175}=2\times5\sqrt{7}=10\sqrt{7}\). Therefore, \(4\sqrt{7}+10\sqrt{7}=14\sqrt{7}\), making option B correct. Exam tip: after reducing surds to the same radical, add or subtract their coefficients only.

\(\sqrt{48}=\sqrt{16\times3}=4\sqrt{3}\), so \(5\sqrt{48}=20\sqrt{3}\). Similarly, \(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\), so \(3\sqrt{75}=15\sqrt{3}\). Therefore, \(20\sqrt{3}-15\sqrt{3}=5\sqrt{3}\), making option A correct. In such questions, first simplify each surd and then combine only like surd terms.

The governing concept is the product rule for square roots: √a × √b = √(ab) for non-negative a and b. Applying it gives √32 × √50 = √(32 × 50) = √1600. Since 1600 = 40², its principal square root is 40. A second verification is possible by simplifying each radical: √32 = √(16×2) = 4√2 and √50 = √(25×2) = 5√2. Their product is (4√2)(5√2) = 20×2 = 40. Thus option A is correct. Option B doubles the answer, option C is an incomplete or incorrect simplification because the two √2 factors must also be multiplied, and option D adds radicands instead of multiplying them.

( \frac{\sqrt{180}}{\sqrt{20}}=\sqrt{9}=3 ). In division of square roots, divide the radicands.

\(\frac{\sqrt{200}}{\sqrt{2}}=\sqrt{\frac{200}{2}}=\sqrt{100}=10\), and \(\sqrt{16}=4\). Therefore, the value is \(10+4=14\). Option B incorrectly reflects a subtraction-based calculation rather than evaluating the quotient and then adding \(\sqrt{16}\). In the exam, remember that for positive radicands, \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\).

Since \(4.84=(2.2)^2\), we have \(\sqrt{4.84}=2.2\). Similarly, \(1.44=(1.2)^2\), so \(\sqrt{1.44}=1.2\). Therefore, the required value is \(2.2-1.2=1\). Options B and D are only the values of the second and first square roots respectively, while option C is not the correct difference. Exam tip: Recognise decimal numbers as squares of terminating decimals before subtracting.

The first term is ( \frac{4}{5} ) and the second is ( \frac{3}{8} ). The sum is ( \frac{32}{40}+\frac{15}{40}=\frac{47}{40} ).