Question

Tính a³+b³ biết a+b=3 và a²+b²=18

Answer 1

Có: \(\left(a+b\right)^2=9\)

=> \(a^2+2ab+b^2=9\)

=>\(2ab=9-\left(a^2+b^2\right)=9-18=-9\)

\(\Rightarrow ab=-4,5\)

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=3\cdot\left(18+4,5\right)=67,5\)

Answer 2

ta có : \(a^2+b^2=18\Leftrightarrow\left(a+b\right)^2-2ab=18\Leftrightarrow3^2-2ab=18\)

\(\Leftrightarrow9-2ab=18\Leftrightarrow2ab=-9\Leftrightarrow ab=\dfrac{-9}{2}\)

ta lại có : \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=3^3-3\left(\dfrac{-9}{2}\right)\left(3\right)\)

\(=27+\dfrac{81}{2}=\dfrac{135}{2}\)