Question

please answer both of them​

Answer 1

Corrected question: To prove that:

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$\sf \dashrightarrow \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}} = 2$

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Proof:

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Let's try to rationalize the denominator of each fraction to see if we'll be able to simplify it. In order to rationalize the denominator, we'll have to multiply both the numerator and the denominator of the fraction by the conjugate of the denominator.

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[Conjugating an expression with two terms basically changes the sign between the two terms from a + to -, or vice-versa.]

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• Conjugate of 1 + √2 is 1 - √2

• Conjugate of √2 + √3 = √2 - √3

• Conjugate of √3 + √4 = √3 - √4

• Conjugate of √8 + √9 = √8 - √9

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$\sf \dashrightarrow \dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + {\dots\dots\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}} = 2$

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$\sf \dashrightarrow \dfrac{1}{1 + \sqrt{2}} \times \dfrac{1 -\sqrt{2}}{1 -\sqrt{2}} \ + \ \dfrac{1}{\sqrt{2} + \sqrt{3}} \times \dfrac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\ \\ \\ {\ \ \ \ \ \ \ \ \ + \ \dfrac{1}{\sqrt{3} + \sqrt{4}} \times \dfrac{\sqrt{3} - \sqrt{4}}{\sqrt{3} - \sqrt{4}} \ + \ {\dots} \ + \dfrac{1}{\sqrt{8} + \sqrt{9}} \times \dfrac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}}= 2}$

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[Broken into two lines for readability]

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Using (a + b)(a - b) = a² - b² we get;

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$\sf \dashrightarrow \ \dfrac{1 -\sqrt{2}}{(1)^2 - (\sqrt{2})^2} + \dfrac{\sqrt{2} - \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} + \dfrac{\sqrt{3} - \sqrt{4}}{(\sqrt{3})^2 - (\sqrt{4})^2} + \ ... \ + \ \dfrac{\sqrt{8} - \sqrt{9}}{(\sqrt{8})^2 - (\sqrt{9})^2} = 2$

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$\sf \dashrightarrow \dfrac{1 -\sqrt{2}}{1 - 2} \ + \ \dfrac{\sqrt{2} - \sqrt{3}}{2 - 3} \ + \ \dfrac{\sqrt{3} - \sqrt{4}}{3 - 4} + \ {\dots} \ + \ \dfrac{\sqrt{8} - \sqrt{9}}{8 - 9} = 2$

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$\sf \dashrightarrow \dfrac{1 -\sqrt{2}}{-1} \ + \ \dfrac{\sqrt{2} - \sqrt{3}}{-1} \ + \ \dfrac{\sqrt{3} - \sqrt{4}}{-1} + \ {\dots} \ + \ \dfrac{\sqrt{8} - \sqrt{9}}{-1} = 2$

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$\sf \dashrightarrow \ \dfrac{1 -\sqrt{2} + \sqrt{2} - \sqrt{3} + \sqrt{3} - \sqrt{4} + \dots + \sqrt{8} - \sqrt{9}}{-1} = 2$

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Everything except for 1 and -√9 gets cancelled.

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$\sf \dashrightarrow \ \dfrac{1 - \sqrt{9}}{-1} = 2$

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$\sf \dashrightarrow \ \dfrac{1 - 3}{-1} = 2$

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$\sf \dashrightarrow \ \dfrac{-2}{-1} = 2$

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$\sf \dashrightarrow \ 2 = 2$

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LHS = RHS

Hence proved.

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Question 2: We're asked to find the values of 'a' and 'b' for the equation;

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$\sf \dashrightarrow \ \dfrac{3 + 2\sqrt{2}}{2 - \sqrt{2}} - \dfrac{3 - 2\sqrt{2}}{2 + \sqrt{2}} = a + b\sqrt{2}$

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On rationalizing the denominator we get;

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$\sf \dashrightarrow \ \dfrac{3 + 2\sqrt{2}}{2 - \sqrt{2}} \times \dfrac{2 + \sqrt{2}}{2 + \sqrt{2}} - \dfrac{3 - 2\sqrt{2}}{2 + \sqrt{2}} \times \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}}= a + b\sqrt{2}$

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Using (a - b)(a + b) = a² - b² in the denominators of both fractions we get;

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$\sf \dashrightarrow \ \dfrac{(3 + 2\sqrt{2})(2 + \sqrt{2})}{(2)^2 - (\sqrt{2})^2} - \dfrac{(3 - 2\sqrt{2})(2 - \sqrt{2})}{(2)^2 - (\sqrt{2})^2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ \dfrac{6 + 3\sqrt{2} + 4\sqrt{2} + 2(\sqrt{2})^2}{4 - 2} - \dfrac{6 - 3\sqrt{2} - 4\sqrt{2} + 2(\sqrt{2})^2}{4 - 2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ \dfrac{6 + 3\sqrt{2} + 4\sqrt{2} + 4}{2} - \dfrac{6 - 3\sqrt{2} - 4\sqrt{2} + 4}{2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ \dfrac{6 + 3\sqrt{2} + 4\sqrt{2} + 4 - (6 - 3\sqrt{2} - 4\sqrt{2} + 4)}{2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ \dfrac{6 + 3\sqrt{2} + 4\sqrt{2} + 4 - 6 + 3\sqrt{2} + 4\sqrt{2} - 4}{2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ \dfrac{3\sqrt{2} + 4\sqrt{2} + 3\sqrt{2} + 4\sqrt{2}}{2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ \dfrac{14\sqrt{2}}{2} = a + b\sqrt{2}$

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$\sf \dashrightarrow \ 7\sqrt{2} = a + b\sqrt{2}$

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∴ a = 0 and b = 7.