Math, asked by prithviawana27, 10 months ago

please help me in doing only first question from each;its from rationalisation see the attachment

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Answered by 1KingArjun
1

Answer:

Q.4 (i) √3 - √2/√3 + √2

= √3 - √2/√3 + √2 × √3 - √2/√3 - √2

= (√3 - √2)(√3 - √2)/(√3 + √2)(√3 - √2)

= (√3)² + (√2)² - 2(√3)(√2) / (√3)² - (√2)²

= 3 + 2 - 2√6 / 3 - 2

= 5 - 2√6

Q.5 (i) (3√2 - 2√3 / 3√2 + 2√3) + (√12/√3 - √2)

= (3√2 - 2√3/3√2 + 2√3 × 3√2 - 2√3/3√2 - 2√3) + (2√3/√3 - √2 × √3 + √2/√3 + √2)

= [(3√2 - 2√3)² / (3√2 + 2√3)(3√2 - 2√3)] + [(2√3)(√3 + √2) / (√3 - √2)(√3 + √2)

= [(3√2)² + (2√3)² - 2(3√2)(2√3) / (3√2)² - (2√3)²] + [6 + 2√6 / (√3)² - (√2)²]

= (18 + 12 - 12√6 / 18 - 12) + (6 + 2√6 / 3 - 2)

= 30 - 12√6 / 6  +  6 + 2√6

= 5 - 2/6 + 6 + 2/6

= 11

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