Question

Cho \(x+\dfrac{1}{x}=a\). Tính các biểu thức sau theo \(a\):

a) \(x^2+\dfrac{1}{x^2}\)

b) \(x^3+\dfrac{1}{x^3}\)

c) \(x^4+\dfrac{1}{x^4}\)

d) \(x^5+\dfrac{1}{x^5}\)

Answer 1

a) ta có \(x+\dfrac{1}{x}=a\Leftrightarrow x^2+\dfrac{1}{x^2}+2=a^2\Leftrightarrow\dfrac{1}{x^2}+x^2=a^2-2\)

Answer 2

b)ta có \(x+\dfrac{1}{x}=a\Leftrightarrow x^3+3x^2.\dfrac{1}{x}+3x\dfrac{1}{x^2}+\dfrac{1}{x^3}=a^3\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}=a^3-3x-3\dfrac{1}{x}=a^3-3a\)

Answer 3

c)ta có \(x^2+\dfrac{1}{x^2}=a^2-2\Leftrightarrow x^4+2x^2\dfrac{1}{x^2}+\dfrac{1}{x^4}=a^4-4a+4\)

\(\Leftrightarrow x^4+\dfrac{1}{x^4}=a^4-4a+2\)

Answer 4

đ) ta có \(\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}=a^2-2\\x^3+\dfrac{1}{x^3}=a^3-3a\end{matrix}\right.\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=\left(a^2-2\right)\left(a^3-3a\right)\)

\(\Leftrightarrow x^5+\dfrac{1}{x}+x+\dfrac{1}{x^5}=a^5-3a^3-2a^3+6a\)

\(\Leftrightarrow x^5+a+\dfrac{1}{x^5}=a^5-5a^3+6a\)

\(\Leftrightarrow x^5+\dfrac{1}{x^5}=a^5-5a^3+5a\)