2021 AMC 10B Spring 第 15 题

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15.

实数 xx 满足方程 求 x117x7+x3x^{11}-7x^{7}+x^3 的值。 x+1x=5.x+\frac{1}{x} = \sqrt{5}.

The real number xx satisfies the equation x+1x=5.x+\frac{1}{x} = \sqrt{5}. What is the value of x117x7+x3?x^{11}-7x^{7}+x^3?

1-1

00

11

22

5\sqrt{5}

答案:B
知识点:代数变形对称性(代数)
难度评级:1340
视频讲解:
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文字解答:

x+1x=5x+\frac{1}{x} = \sqrt{5},平方得 再平方一次: 即 两边乘以 x7x^7,得到 x2+2+1x2=5x^2 + 2 + \frac 1{x^2} = 5 x2+1x2=3.x^2 + \frac 1{x^2} = 3. x4+2+1x4=9x^4 + 2 + \frac 1{x^4} = 9 x47+1x4=0.x^4 -7 + \frac 1{x^4} = 0. x117x7+x3=0.x^{11}-7x^{7}+x^3=0.

所以答案是 B

Since x+1x=5,x+\frac{1}{x} = \sqrt{5}, squaring yields x2+2+1x2=5x^2 + 2 + \frac 1{x^2} = 5 x2+1x2=3.x^2 + \frac 1{x^2} = 3. Squaring again yields x4+2+1x4=9x^4 + 2 + \frac 1{x^4} = 9 x47+1x4=0.x^4 -7 + \frac 1{x^4} = 0. Multiplying by x7x^7 yields x117x7+x3=0.x^{11}-7x^{7}+x^3=0.

Thus, the answer is B .

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