Real Numbers

Use the identity \((a+b)^2-(a-b)^2=4ab\). Here, \(a=\sqrt{7}\) and \(b=\sqrt{2}\), so the value is \(4\times\sqrt{7}\times\sqrt{2}=4\sqrt{14}\). \(2\sqrt{14}\) results from incorrectly using 2 instead of 4. Exam tip: Apply this identity directly instead of expanding both squares.

( \sqrt{20}=2\sqrt{5} ) and ( \sqrt{45}=3\sqrt{5} ). Dividing (5\sqrt{5}) by ( \sqrt{5} ) gives (5).

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. Hence, the claim that the sum of two irrational numbers is always irrational is false. In option B, \(\sqrt{2}+\sqrt{3}\) is still irrational, so it is not a counterexample. Exam tip: To disprove a statement containing words such as “always,” it is enough to find one valid counterexample.

( (\sqrt{10}-2)^2=10-4\sqrt{10}+4=14-4\sqrt{10} ). The positive principal square root is ( \sqrt{10}-2 ).

In \(0.101001000100001\ldots\), the number of zeros between successive 1s keeps increasing. Hence, no fixed block of digits repeats, so its decimal expansion is non-terminating and non-recurring; therefore, it is irrational. \(\frac{7}{125}\) has a terminating decimal expansion, while \(0.\overline{27}\) is recurring. Also, \(\sqrt{81}=9\), which is rational. Exam tip: A non-terminating decimal is irrational only when it does not repeat a fixed pattern.

When \(p/q\) is in lowest terms, its decimal expansion terminates exactly when the denominator \(q\) has only 2 and 5 as prime factors, that is, \(q=2^m5^n\). If the denominator has any other prime factor, the decimal expansion is non-terminating recurring, but the number is still rational. Hence options C and D are incorrect. Exam tip: Always reduce a fraction to lowest terms before checking its decimal expansion.

( \sqrt{80}\approx8.94 ), so ( -\sqrt{80}\approx-8.94 ). It is greater than ( -9 ) because it is closer to zero.

( \frac{96}{384}=\frac{1}{4} ) and (4=2^2). A decimal terminates when the denominator has only (2) and (5) as prime factors.

( \frac{132}{360}=\frac{11}{30} ), and the denominator also has (3). Therefore it gives a non-terminating repeating decimal.

The block (81) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.

\(\sqrt{28}=\sqrt{4\times7}=2\sqrt{7}\), \(\sqrt{63}=\sqrt{9\times7}=3\sqrt{7}\), and \(\sqrt{112}=\sqrt{16\times7}=4\sqrt{7}\). Hence, the sum is \(\sqrt{7}+2\sqrt{7}+3\sqrt{7}+4\sqrt{7}=(1+2+3+4)\sqrt{7}=10\sqrt{7}\). An option such as \(8\sqrt{7}\) can result from using an incorrect coefficient for one term. Exam tip: extract perfect-square factors from surds before adding like surd terms.

Use a common denominator for the two fractions. The denominator product is \\( (\sqrt7-\sqrt6)(\sqrt7+\sqrt6)=7-6=1\\), by the difference-of-squares identity. The combined numerator is \\(\sqrt7+\sqrt6-(\sqrt7-\sqrt6)=2\sqrt6\\). Since the denominator equals 1, the whole expression is \\(2\sqrt6\\). Therefore option A is correct.

The subtraction sign is important: subtracting the second numerator changes both signs inside its parentheses, so the \\(\sqrt7\\) terms cancel and the two \\(\sqrt6\\) terms add. Both denominators are nonzero because \\(\sqrt7\ne\sqrt6\\). A numerical check gives a value near 4.90, matching \\(2\sqrt6\\). Thus the supplied answer A is valid and follows directly from conjugates and the difference-of-squares formula.

Use the addition rule for two fractions and the conjugate identity \((a-b)(a+b)=a^2-b^2\). The common denominator is \((5-\sqrt{21})(5+\sqrt{21})=25-21=4\). The numerator is \((5+\sqrt{21})+(5-\sqrt{21})=10\), because the irrational terms cancel. Therefore the expression equals \(10/4=5/2\), so option A is correct. Option B incorrectly uses 21 as the effective denominator, option C ignores the cancellation, and option D misses the factor of 2 produced by the two fractions. The denominator is nonzero, so the calculation is valid.

This is a product of conjugates in the form \((a+b)(a-b)=a^2-b^2\). Here, \(a=\sqrt{10}\) and \(b=\sqrt{3}\), so the value is \(10-3=7\). \(2\sqrt{30}\) relates to the middle terms in expansion, but these terms cancel for conjugates. Exam tip: Whenever you see \((x+y)(x-y)\), apply the difference of squares formula directly.

They become (2\sqrt{5},6\sqrt{5},4\sqrt{5},3\sqrt{5}). Since (2+6-4-3=1), the result should be ( \sqrt{5} ).

The governing method is rationalisation using the conjugate of the denominator. Starting from 1/x = 1/(4 − √7), multiply numerator and denominator by 4 + √7. The denominator becomes (4 − √7)(4 + √7) = 4² − (√7)² = 16 − 7 = 9, while the numerator is 4 + √7. Thus 1/x = (4 + √7)/9, so option A is correct. Option B gives the numerator but omits division by 9. Option C incorrectly keeps the original sign instead of using the conjugate. Option D multiplies by 9 rather than dividing by the rationalised denominator. Multiplying the proposed answer by x gives [(4 + √7)(4 − √7)]/9 = 9/9 = 1.

(a^2=7+2\sqrt{10}) and ( \frac{1}{a^2} ) is not simply (7-2\sqrt{10}) because (a(\sqrt{5}-\sqrt{2})=3). Check options carefully using reciprocal rules.

(x-3=\sqrt{8}=2\sqrt{2}), so ( \frac{1}{x-3}=\frac{\sqrt{2}}{4} ). The total should be (3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}).

\(\sqrt{x^2}=|x|\) because the principal square root is always non-negative. Taking \(x=a-b\), we get \(\sqrt{(a-b)^2}=|a-b|\). The expression \(a-b\) is correct only when \(a\ge b\); if \(a<b\), it is negative. Exam tip: whenever a squared expression comes out of a square root, remember to use absolute value signs.

\(\sqrt{x^2}=|x|\), because the principal square root is always non-negative. Thus, for \(x=-9\), \(\sqrt{x^2}=|-9|=9\) and \(3x=3(-9)=-27\). Therefore, \(\sqrt{x^2}+3x=9-27=-18\). Taking \(\sqrt{x^2}\) directly as \(x\) would give \(-36\), which is incorrect. Exam tip: always rewrite \(\sqrt{x^2}\) as \(|x|\).