AMC8 2022

AMC8 2022 · Q2

AMC8 2022 · Q2. It mainly tests Algebra misc.

Consider these two operations: \begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*} What is the output of $(5 \, \blacklozenge \, 3) \, \bigstar \, 6?$

考虑以下两个运算： \begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*} 什么是 $(5 \, \blacklozenge \, 3) \, \bigstar \, 6$ 的输出？

(A) {-}20 {-}20

(B) 4 4

(D) 100 100

(E) 220 220

Answer

Correct choice: (D)

正确答案：（D）

Solution

We can find a general solution to any $((a \, \blacklozenge \, b) \, \bigstar \, c)$. \[((a \, \blacklozenge \, b) \, \bigstar \, c)\] \[=((a^2-b^2) \, \bigstar \, c)\] \[=(a^2-b^2-c)^2\] \[=a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2\] \[=5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2\] \[=625+81-225-300-225+108+36\] \[=\boxed{\textbf{(D) } 100}\] To time wasting

我们可以找到任何 $((a \, \blacklozenge \, b) \, \bigstar \, c)$ 的一般解。 \[((a \, \blacklozenge \, b) \, \bigstar \, c)\] \[=((a^2-b^2) \, \bigstar \, c)\] \[=(a^2-b^2-c)^2\] \[=a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2\] \[=5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2\] \[=625+81-225-300-225+108+36\] \[=\boxed{\textbf{(D) } 100}\]

Topics

AL.020 Algebra misc