AMC10 2021 B

AMC10 2021 B · Q15

AMC10 2021 B · Q15. It mainly tests Exponents & radicals, Manipulating equations.

The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$. What is the value of $x^{11}-7x^{7}+x^3?$

实数$x$满足方程$x+\frac{1}{x} = \sqrt{5}$。求$x^{11}-7x^{7}+x^3$的值？

(A) -1 -1

(B) 0 0

(D) 2 2

(E) \sqrt{5} \sqrt{5}

Answer

Correct choice: (B)

正确答案：（B）

Solution

We square $x+\frac{1}{x}=\sqrt5$ to get $x^2+2+\frac{1}{x^2}=5$. We subtract 2 on both sides for $x^2+\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\frac{1}{x^4}=9$ so $x^4+\frac{1}{x^4}=7$. We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x^4-7+\frac{1}{x^4})$. Therefore because $x^4+\frac{1}{x^4}$ is 7, it is equal to $x^7(0)=\boxed{\textbf{(B) } 0}$.

平方$x+\frac{1}{x}=\sqrt{5}$得$x^2 + 2 + \frac{1}{x^2} = 5$，即$x^2 + \frac{1}{x^2} = 3$。再平方得$x^4 + 2 + \frac{1}{x^4} = 9$，即$x^4 + \frac{1}{x^4} = 7$。原式$x^{11} - 7x^7 + x^3 = x^7(x^4 - 7 + \frac{1}{x^4}) = x^7(7 - 7) = x^7 \cdot 0 = \boxed{\textbf{(B) } 0}$。

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