Question

1. a) Cho x²+y²=10; xy=2

Tính x+y; x-y; x²-y²

b) Cho x+y=1

Tính A=x³+y³+3xy

Hãy giúp mính nhé.

 

Answer 1

\(x^2+y^2+2xy\)

\(=10+2.2\)

\(=14\)

\(\Rightarrow\left(x+y\right)^2=14\)

\(\Rightarrow x+y=\sqrt{14}\)

Answer 2

(x-y)² = x² - 2xy +y² = 10 -4 = 6

x-y =\(\sqrt{6}\)

x² -y² =(x+y)(x-y) = \(\sqrt{14.6}\)= \(\sqrt{84}\)