[toan 8] bien doi dong nhat .giup voi...

B

braga

$\left(x+\dfrac{1}{x}\right)^2=x^2+\dfrac{1}{x^2}-2=9\implies x+\dfrac{1}{x}=3 \\ \text{Ta có:} \ x^3+\dfrac{1}{x^3}=\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)=7.3-3=18 \\ x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=49-2=47\\ \implies x^5+\dfrac{1}{x^5}=\left(x^4+\dfrac{1}{x^4}\right)\left(x+\dfrac{1}{x}\right)-\left(x^3+\dfrac{1}{x^3}\right)=47.3-18=\boxed{123}$
 
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H

hoangcoi9999

Co:
[TEX](\frac{1}{x^2}+x^2)(\frac{1}{x^3}+x^3)= \frac{1}{x^5}+x^5+\frac{1}{x}+x \Rightarrow \frac{1}{x^5}+x^5 =(\frac{1}{x^2}+x^2)(\frac{1}{x^3}+x^3)- (\frac{1}{x}+x) = 9.18-3=123 [/TEX]
 
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