AMC10 2020 A

AMC10 2020 A · Q14

AMC10 2020 A · Q14. It mainly tests Quadratic equations, Manipulating equations.

Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$. What is the value of $\frac{x + x^{3}}{y^{2} + y^{3}} \times (x^{2} + y)$?

实数 $x$ 和 $y$ 满足 $x + y = 4$ 和 $x \cdot y = -2$。求 $\frac{x + x^{3}}{y^{2} + y^{3}} \times (x^{2} + y)$ 的值？

(A) 360 360

(B) 400 400

(D) 440 440

(E) 480 480

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The given expression equals $$ \frac{x^3y^2+x^5+y^5+y^3x^2}{x^2y^2}=\frac{(x^2+y^2)(x^3+y^3)}{(xy)^2}. $$ Note that $$ x^2+y^2=(x+y)^2-2xy=4^2-2(-2)=20. $$ Because $(x+y)^3=x^3+y^3+3x^2y+3xy^2$, it follows that $$ \begin{aligned} x^3+y^3&=(x+y)^3-3x^2y-3xy^2\\ &=(x+y)^3-3xy(x+y)\\ &=4^3-3(-2)(4)\\ &=88. \end{aligned} $$ The requested value is $$ \frac{20\cdot 88}{(-2)^2}=440. $$

答案（D）：给定的式子等于 $$ \frac{x^3y^2+x^5+y^5+y^3x^2}{x^2y^2}=\frac{(x^2+y^2)(x^3+y^3)}{(xy)^2}. $$ 注意 $$ x^2+y^2=(x+y)^2-2xy=4^2-2(-2)=20. $$ 因为 $(x+y)^3=x^3+y^3+3x^2y+3xy^2$，所以 $$ \begin{aligned} x^3+y^3&=(x+y)^3-3x^2y-3xy^2\\ &=(x+y)^3-3xy(x+y)\\ &=4^3-3(-2)(4)\\ &=88. \end{aligned} $$ 所求的值为 $$ \frac{20\cdot 88}{(-2)^2}=440. $$

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