toán lớp 8

C

chonhoi110

cho (x^n-x^-n)/(x^n+x^-n) =m với n thuộc N*.tính P=(x^2n-x^-2n)/(x^2n+x^-2n)
Ta có: $1+m=\dfrac{2x^n}{x^n+\dfrac{1}{x^n}} ; 1-m=\dfrac{2}{x^n(x^n+\dfrac{1}{x^n})} \rightarrow \dfrac{1+m}{1-m}=x^{2n}$

$\rightarrow \dfrac{1-m}{1+m}=x^{-2n}$

Do đó $P=\dfrac{\dfrac{1+m}{1-m}-\dfrac{1-m}{1+m}}{\dfrac{1+m}{1-m}+\dfrac{1-m}{1+m}}=\dfrac{(1+m)^2-(1-m)^2}{(1-m)(1+m)}.\dfrac{(1-m)(1+m)}{(1+m)^2+(1-m)^2}=\dfrac{2m}{1+m^2}$
 
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