Bài này cần có công thức:
Ta có:\(x+\frac{1}{x}=3=>x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=9-2=7\)
Lại có: \(x^5+\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)\)
=\(7\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)-3=7.3.6-3=123\)
Vậy \(x^5+\frac{1}{x^5}=123\)
con này không nhầm có lời giửi rồi!
\(\left(x+\frac{1}{x}\right)=3\Rightarrow x^2+\frac{1}{x^2}=7\Rightarrow x^4+\frac{1}{x^4}=47\)
\(3.7=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=\left(x^3+\frac{1}{x^3}\right)+\left(x+\frac{1}{x}\right)\)
\(\Rightarrow\left(x^3+\frac{1}{x^3}\right)=3.7-3=3.6\)
\(3.47=\left(x+\frac{1}{x}\right)\left(x^4+\frac{1}{x^4}\right)=\left(x^5+\frac{1}{x^5}\right)+x^3+\frac{1}{x^3}\\ \)
\(x^5+\frac{1}{x^5}=3.47-3.6=3\left(47-6\right)=3.41=123\)
Ta có: \(x+\dfrac{1}{x}=3\Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\Rightarrow x^2+\dfrac{1}{x^2}+2=9\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=7\)
Lại có: \(x+\dfrac{1}{x}=3\Rightarrow\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\)
\(\Rightarrow x^3+\dfrac{1}{x^3}+9=27\Rightarrow x^3+\dfrac{1}{x^3}=18\)
Do đó: \(\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=x^5+\dfrac{1}{x^5}+x+\dfrac{1}{x}=7.18=126\)
\(\Rightarrow x^5+\dfrac{1}{x^5}+3=126\Rightarrow x^5+\dfrac{1}{x^5}=123\)
Vậy \(x^5+\dfrac{1}{x^5}=123\)
Nhanh nek:Phân tích
\(\left(x+\dfrac{1}{x}\right)^5=x^5+5x^3+10x+10.\dfrac{1}{x}+5\dfrac{1}{x^3}+\dfrac{1}{x^5}\)
\(=x^5+\dfrac{1}{x^5}+5\left(x^3+\dfrac{1}{x^3}\right)+10\left(x+\dfrac{1}{x}\right)=243\)
\(\left(x+\dfrac{1}{x}\right)^3=x^3+3x+\dfrac{3}{x}+\dfrac{1}{x^3}=x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)\)
\(=x^3+\dfrac{1}{x^3}+9=27\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
Thay các số liệu vào phương trình đầu bài:
\(x^5+\dfrac{1}{x^5}+5.18+10.3=243\)
\(\Leftrightarrow x^5+\dfrac{1}{x^5}=123\)
