Question

X2+1/x2 when 2+1/x = 20​

Answer 1

Correct Question :-

$\sf Find \:{x}^{2} + \dfrac{1}{ {x}^{2} } \:when \:x + \dfrac{1}{x} = 20$

Given :-

→ $\sf x + \dfrac{1}{x} = 20$

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To Find :-

→ $\sf {x}^{2} + \dfrac{1}{ {x}^{2} }$

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Solution :-

$\sf x + \dfrac{1}{x} = 20$

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Squaring on both side -

→ $\sf (x + \dfrac{1}{x})^2= (20)^2$

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→ $\sf x^2 + \dfrac{1}{x^2} + 2 \times \cancel x \times \frac{1}{\cancel x} = 400$

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→ $\sf x^2 + \dfrac{1}{x^2} + 2 = 400$

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→ $\sf {x}^{2} + \dfrac{1}{ {x}^{2} } = 400 - 2$

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→ $\sf {x}^{2} + \dfrac{1}{ {x}^{2} } = 398$

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Hence the value of $\boxed{\sf {x}^{2} + \dfrac{1}{ {x}^{2} } = 398}$

Answer 2

(1+x2)40⋅(x2+2+x21)−5

=(1+x2)40⋅[(x+x1)2]−5

=(1+x2)40⋅[x2(1+x2)2]−5

=(1+x2)40⋅[x−10(1+x2)−10]

=(1+x2)30⋅x10

⇒ Coefficient of x20 in (1+x2)30⋅x10 is 30C5or 30C