Question

Cho x+y=2; x²+y²=10 . Tính x³+y³.

Answer 1

\(x+y=2\\ \Rightarrow\left(x+y\right)^2=4\\ \Rightarrow x^2+2xy+y^2=4\\ \Rightarrow10+2xy=4\\ \Rightarrow2xy=-6\\ \Rightarrow xy=-3\)

Ta có :

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10+3\right)=2\cdot13=26\)

=> KL : ....

Answer 2

x + y = 2

<=> y = 2 - x

x² + y² = 10

<=> x² + (2 - x)² = 10

<=> x² + x² - 4x + 4 - 10 = 0

<=> 2x² - 4x - 6 = 0

<=> 2x² + 2x - 6x - 6 = 0

<=> 2x.(x + 1) - 6.(x + 1)

<=> (2x - 6)(x + 1) = 0

<=> 2x - 6 = 0 hoặc x + 1 =0

<=> x = 3 hoặc x = -1 <=> y = -1 hoặc y = 3

<=> x³ + y³ = 3³ + (-1³) = 24