Question

Cho x khác 0 và \(x+\dfrac{1}{x}=a\) là hằng số. Tính:

A= \(x^2+\dfrac{1}{x^2}\)

B= \(x^3+\dfrac{1}{x^3}\)

C= \(x^6+\dfrac{1}{x^6}\)

D= \(x^7+\dfrac{1}{x^7}\)

Answer 1

Lời giải:

\(A=x^2+\frac{1}{x^2}=x^2+2.x.\frac{1}{x}+\frac{1}{x^2}-2\)

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

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\(B=x^3+\frac{1}{x^3}=(x^2+\frac{1}{x^2})(x+\frac{1}{x})-x-\frac{1}{x}\)

\(=(x^2+\frac{1}{x^2})(x+\frac{1}{x})-(x+\frac{1}{x})=(a^2-2).a-a=a^3-3a\)

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\(C=x^6+\frac{1}{x^6}=(x^3+\frac{1}{x^3})^2-2.x^3.\frac{1}{x^3}\)

\(=(a^3-3a)^2-2\)

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\(D=x^7+\frac{1}{x^7}=(x^6+\frac{1}{x^6})(x+\frac{1}{x})-x^5-\frac{1}{x^5}\)

\(=(x^6+\frac{1}{x^6})(x+\frac{1}{x})-[(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})-(x+\frac{1}{x})]\)

\(=[(a^3-3a)^2-2]a-[(a^2-2)(a^3-3a)-a]\)

\(=a^7+14a^3-7a^5-7a\)