Question

solve the answer
X+1/X=2 then x^2020+1/x^2020=?

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Answer 1

Given,

$\longrightarrow x+\dfrac{1}{x}=2$

$\longrightarrow \dfrac{x^2+1}{x}=2$

Multiplying by $x,$

$\longrightarrow x^2+1=2x$

Subtracting $2x,$

$\longrightarrow x^2-2x+1=0$

Factorising LHS we get,

$\longrightarrow(x-1)^2=0$

This implies,

$\longrightarrow x=1$

Therefore,

$\longrightarrow x^{2020}+\dfrac{1}{x^{2020}}=1^{2020}+\dfrac{1}{1^{2020}}$

$\longrightarrow x^{2020}+\dfrac{1}{x^{2020}}=1+\dfrac{1}{1}$

$\longrightarrow x^{2020}+\dfrac{1}{x^{2020}}=1+1$

$\longrightarrow\underline{\underline{x^{2020}+\dfrac{1}{x^{2020}}=2}}$

Hence 2 is the answer.

Answer 2

Answer:-

$\sf x + \dfrac{1}{x} = 2$

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

$\longrightarrow \sf {x}^{2} + 1 = 2x$

$\longrightarrow \sf {x}^{2} - 2x + 1 = 0$

$\longrightarrow \sf {x}^{2} - x - x + 1 = 0$

$\longrightarrow \sf x(x - 1) - 1(x - 1) = 0$

$\longrightarrow \sf(x - 1)(x - 1) = 0$

So,

$\longrightarrow \sf x = x_1 = x_2 = 1$

Hence,

$\longrightarrow \sf {x}^{2020 } + \dfrac{1}{ {x}^{2020} }$

$\longrightarrow \sf {1}^{2020} + \dfrac{1}{ {1}^{2020} }$

$\longrightarrow \sf 1 + \dfrac{1}{1}$

$\longrightarrow \sf 1 + 1$

$\longrightarrow \underline{\underline{\sf{\green{\quad 2 \quad}}}}$