Question

定义 $a@b = ab - b^2$ 和 $a\#b = a + b - ab^2$ 。求下式的值： $\dfrac{6@2}{6\#2}\text{？}$

Define $a@b = ab - b^2$ and $a\#b = a + b - ab^2.$ What is $\dfrac{6@2}{6\#2}?$

$-\dfrac{1}{2}$

$-\dfrac{1}{4}$

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

将 $a = 6$ 、 $b = 2$ 分别代入两个定义。

Substitute $a = 6$ and $b = 2$ into each definition

分子是 $6 \cdot 2 - 2^2$ ，分母是 $6 + 2 - 6 \cdot 2^2$ 。

The numerator is $6 \cdot 2 - 2^2$ and the denominator is $6 + 2 - 6 \cdot 2^2$

分子为 $6@2 = 6 \cdot 2 - 2^2 = 12 - 4 = 8$ 。

分母为 $6\#2 = 6 + 2 - 6 \cdot 2^2$ $= 8 - 24$ $= -16$ 。

商为 $\dfrac{8}{-16} = -\dfrac{1}{2}\text{。}$

所以正确答案是 A。

The numerator is $6@2 = 6 \cdot 2 - 2^2 = 12 - 4 = 8.$

The denominator is $6\#2 = 6 + 2 - 6 \cdot 2^2$ $= 8 - 24$ $= -16.$

The quotient is $\dfrac{8}{-16} = -\dfrac{1}{2}.$

Thus, the correct answer is A.