Toán nâng cao đại số 8

V

vipboycodon

Ta có: $(x+\dfrac{1}{x})^2 = x^2+\dfrac{1}{x^2}+2 = 9 \rightarrow x+\dfrac{1}{x} = 3$
$x^3+\dfrac{1}{x^3} = (x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-1) = 18$
$x^5+\dfrac{1}{x^5} = (x^2+\dfrac{1}{x^2})(x^3+\dfrac{1}{x^3})-(x+\dfrac{1}{x}) = 123$
 
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