Pembahasan Matematika Dasar SIMAK UI 2016 #1

Soal

Nilai dari \( \sqrt{(100)(101)(102)(103) + 1} = \ldots\)

A. 10101

B. 10201

C. 10301

D. 10401

E. 10501

Jawab

Misalkan \(x = 101\), maka

$$ \begin{align} \sqrt{(x-1)(x)(x+1)(x+2) + 1} &= \sqrt{ (x^2 + x)(x^2 + x - 2) + 1}\\ &= \sqrt{(x^2 + x)^2 - 2(x^2 + x) + 1}\\ &= \sqrt{(x^2 + x -1)^2}\\ &= x^2 + x - 1\\ &= 101^2 + 100 = 10201 + 100 = 10301 \end{align} $$

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