Math, asked by singumanojs, 8 months ago

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

Answers

Answered by shadowsabers03
12

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.

Answered by Arceus02
25

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}}}}

Similar questions