Question

Cho \(x+\frac{1}{x}=a\) Tính giá trị của các biểu thức sau theo a:

a) \(x^4+\frac{1}{x^4}\)                                      b) \(x^5+\frac{1}{x^5}\)

Answer 1

tương tự : 

\(x+\frac{1}{x}=a\)

\(x^5+\frac{1}{x^5}+5x^3+10x+\frac{10}{x}+\frac{5}{x^3}=a^5\)

\(\Rightarrow x^5+\frac{1}{x^5}=a^5-5\left(x^3+\frac{1}{x^3}\right)-10\left(x+\frac{1}{x}\right)\)

Mà : \(x+\frac{1}{x}=a\Rightarrow x^3+\frac{1}{x^3}=a^3-3x-\frac{3}{x}=a^3-3a\)

\(\Rightarrow x^5+\frac{1}{x^5}=a^5-5\left(a^3-3a\right)-10a\)

\(\Rightarrow x^5+\frac{1}{x^5}=a^5-5a^3+15a-10a=a^5-5a^3+5a\)

nha

Answer 2

a) Ta có \(x+\frac{1}{x}=a\)

\(\Rightarrow x^4+4x^2+6+\frac{4}{x^2}+\frac{1}{x^4}=a^4\)

\(\Rightarrow x^4+\frac{1}{x^4}=a^4-6-4\left(x^2+\frac{1}{x^2}\right)\)

Mà \(x+\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)

\(\Rightarrow x^4-\frac{1}{x^4}=a^4-6-4a^2+8=a^4-4a^2+2\)