Question

Answer 1

5: Ta có: \(x+y=3\)

nên \(\left(x+y\right)^2=9\)

\(\Leftrightarrow x^2+y^2+2xy=9\)

\(\Leftrightarrow2xy+25=9\)

hay xy=-8

\(B=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)

\(=3^3-3\cdot\left(-8\right)\cdot3\)

\(=27+9\cdot8=99\)

Answer 2

Lời giải:
5. $xy=\frac{(x+y)^2-(x^2+y^2)}{2}=\frac{3^2-5}{2}=2$

$B=x^3+y^3=(x+y)(x^2-xy+y^2)=3(5-2)=9$

6.

$xy=\frac{x^2+y^2-(x-y)^2}{2}=\frac{15-5^2}{2}=-5$

$C=x^3-y^3=(x-y)(x^2+xy+y^2)=5(15-5)=50$