cho x>o thoả x^2+1/x^2=7.tính
a/ x^3+1/x^3
b/x^5+1/x^5
[TEX]\frac{1}{x^2}+x^2+2=9 \Rightarrow (x+\frac{1}{x})^2=9 \Rightarrow x+\frac{1}{x}=3[/TEX]
\Rightarrow [TEX](x+\frac{1}{x})(x^2+\frac{1}{x^2})=7.3 [/TEX]
\Rightarrow [TEX]x^3+\frac{1}{x^3}+x+\frac{1}{x}=21 \Rightarrow x^3+\frac{1}{x^3}+3=21 \Rightarrow x^3+\frac{1}{x^3}=18[/TEX]
lại có: [TEX](x^3+\frac{1}{x^3})(x^2+\frac{1}{x^2})=18.7[/TEX]
\Rightarrow [TEX]x^5+\frac{1}{x^5}+x+\frac{1}{x}=126[/TEX]
\Rightarrow [TEX]x^5+\frac{1}{x^5}+3=126 \Rightarrow x^5+\frac{1}{x^5}=123[/TEX]