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Solving Inequalities

47. |x+1|/|x-...

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47. |x+1|/|x-1|>1

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_{x^2+1/x^2=47 then x^2-1/x^2=?}

31 x + 47 y = 15 and 47 x + 31 y = 63,

we get:
(a) x = 2, y = −1
(b) x = −2, y = 1
(c) x = 1, y = 2
(d) x = 3, y = −2

x^{4} + \frac{1}{x^{4}} = 47

, then the value of

x^{3} + \frac{1}{x^{3}}

x^4 + 1/x^4 = 47 and x>0 then the value of (x^6 + 1)/x^3 is

(a)18. (b)20. (c)23 . (d)24

(x + 1) (2 x - 1) - 5 x = (2 x - 3) (x - 5) + 47

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