34+7\dfrac{3}{4 + \sqrt{7}}4+73 is equal to :
13(4−7)\dfrac{1}{3}(4 - \sqrt{7})31(4−7)
3(4−7)3(4 - \sqrt{7})3(4−7)
13(4+7)\dfrac{1}{3}(4 + \sqrt{7})31(4+7)
3(4+7)3(4 + \sqrt{7})3(4+7)
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Rationalizing,
⇒34+7×4−74−7⇒3(4−7)(4)2−(7)2⇒3(4−7)16−7⇒3(4−7)9⇒13(4−7).\Rightarrow \dfrac{3}{4 + \sqrt{7}} \times \dfrac{4 - \sqrt{7}}{4 - \sqrt{7}} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{16 - 7} \\[1em] \Rightarrow \dfrac{3(4 - \sqrt{7})}{9} \\[1em] \Rightarrow \dfrac{1}{3}(4 - \sqrt{7}).⇒4+73×4−74−7⇒(4)2−(7)23(4−7)⇒16−73(4−7)⇒93(4−7)⇒31(4−7).
Hence, Option 1 is the correct option.
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227+31243\dfrac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}}43227+312 is equal to :
232\sqrt{3}23
323\sqrt{2}32
3
3+2\sqrt{3} + \sqrt{2}3+2
(5−3)2(\sqrt{5} - \sqrt{3})^2(5−3)2 is :
8+2158 + 2\sqrt{15}8+215
8+158 + \sqrt{15}8+15
8−158 - \sqrt{15}8−15
8−2158 - 2\sqrt{15}8−215
17−5\dfrac{1}{7 - \sqrt{5}}7−51 is equal to :
4(7+5)4(7 + \sqrt{5})4(7+5)
144(7+5)\dfrac{1}{44}(7 + \sqrt{5})441(7+5)
144(7−5)\dfrac{1}{44}(7 - \sqrt{5})441(7−5)
4(7−5)4(7 - \sqrt{5})4(7−5)
If x = 2−1, then (x−1x)2\sqrt{2} - 1, \text{ then } \Big(x - \dfrac{1}{x}\Big)^22−1, then (x−x1)2 is :
222\sqrt{2}22
2
4
2−22 - \sqrt{2}2−2
