If x = 1 + 2, then (x+1x)2\sqrt{2}, \text{ then } \Big(x + \dfrac{1}{x}\Big)^22, then (x+x1)2 is :
222\sqrt{2}22
8
4
424\sqrt{2}42
18 Likes
Given,
x = 1+21 + \sqrt{2}1+2
1x=11+2\dfrac{1}{x} = \dfrac{1}{1 + \sqrt{2}}x1=1+21
Rationalizing,
⇒11+2×1−21−2=1−2(1)2−(2)2=1−21−2=1−2−1=−1+2.⇒(x+1x)2=[1+2+(−1+2)]2=[1−1+2+2]2=[22]2=8.\Rightarrow \dfrac{1}{1 + \sqrt{2}} \times \dfrac{1 - \sqrt{2}}{1 - \sqrt{2}} \\[1em] = \dfrac{1 - \sqrt{2}}{(1)^2 - (\sqrt{2})^2} \\[1em] = \dfrac{1 - \sqrt{2}}{1 - 2} \\[1em] = \dfrac{1 - \sqrt{2}}{-1} \\[1em] = -1 + \sqrt{2}. \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = [1 + \sqrt{2} + (-1 + \sqrt{2})]^2 \\[1em] = [1 - 1 + \sqrt{2} + \sqrt{2}]^2 \\[1em] = [2\sqrt{2}]^2 \\[1em] = 8.⇒1+21×1−21−2=(1)2−(2)21−2=1−21−2=−11−2=−1+2.⇒(x+x1)2=[1+2+(−1+2)]2=[1−1+2+2]2=[22]2=8.
Hence, Option 2 is the correct option.
Answered By
13 Likes
Write a pair of irrational numbers whose product is rational.
If x = 5−2,x+1x\sqrt{5} - 2, x + \dfrac{1}{x}5−2,x+x1 is equal to :
252\sqrt{5}25
454\sqrt{5}45
-4
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
