If x2 + 1x2\dfrac{1}{\text{x}^2}x21 = 3, the value of x - 1x\dfrac{1}{\text{x}}x1 is:
1
-1
±1\pm 1±1
0
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Given, x2 + 1x2\dfrac{1}{\text{x}^2}x21 = 3
As we know
⇒(x−1x)2=x2+1x2−2×x×1x⇒(x−1x)2=x2+1x2−2⇒(x−1x)2=3−2⇒(x−1x)2=1⇒x−1x=1⇒x−1x=±1\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 3 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 1\\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{1}\\[1em] \Rightarrow x - \dfrac{1}{x} = \pm 1⇒(x−x1)2=x2+x21−2×x×x1⇒(x−x1)2=x2+x21−2⇒(x−x1)2=3−2⇒(x−x1)2=1⇒x−x1=1⇒x−x1=±1
Hence, option 3 is the correct option.
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If x2+1x=313\dfrac{x^2 + 1}{x} = 3\dfrac{1}{3}xx2+1=331 and x > 1; find :
(i) x−1xx - \dfrac{1}{x}x−x1
(ii) x3−1x3x^3 - \dfrac{1}{x^3}x3−x31
The difference between two positive numbers is 4 and the difference between their cubes is 316. Find :
(i) their product
(ii) the sum of their squares.
If a3 + b3 + c3 = 3abc, then a + b + c is equal to:
196×196×196+204×204×204(196)2+(204)2−196×204\dfrac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204}(196)2+(204)2−196×204196×196×196+204×204×204 is equal to:
400
-8
8
none of these
