Question

Cho x>0 thỏa mãn x²+\(\frac{1}{x^2}\)=23.Tính giá trị biểu thức : x⁵+\(\frac{1}{x^5}\).

Answer 1

ta có \(x^2+\frac{1}{x^2}\)

=\(\left(x+\frac{1}{x}\right)^2-2x\frac{1}{x}=\left(x+\frac{1}{x}\right)^2-2\)

=> \(\left(x+\frac{1}{x}\right)^2=25.vì\)\(x>0\Rightarrow x+\frac{1}{x}>0\Rightarrow x+\frac{1}{x}=5\)

\(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}=x^3+\frac{1}{x^3}+15\)

\(\Rightarrow x^3+\frac{1}{x^3}=5^3+15=110\)

\(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+x+\frac{1}{x}=x^5+\frac{1}{x^5}+5\)

\(\Rightarrow x^5+\frac{1}{x^5}=23\cdot110-5=2525\)

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