Question

Cho a+b> 8 và \(b\ge3\) . cm :27a²+10b³ >945

Answer 1

Đặt \(\left\{{}\begin{matrix}a+b=8+x\\b=3+y\end{matrix}\right.\left(x,y\in N,xy\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5+x-y\\b=3+y\end{matrix}\right.\)

Khi đó:

\(27a^2+10b^3=27\left(5+x-y\right)^2+10\left(3+y\right)^3\)

\(=27\left(25+x^2+y^2+10x-2xy-10y\right)+10\left(27+y^3+9y^2+27y\right)\)

\(=945+27\left(x^2+y^2-2xy\right)+270x+10y^3+90y^{2\text{​​}}\)

\(=945+27\left(x-y\right)^2+270x+10y^3+90y^2>945\)

Vậy \(27a^2+10b^3>945\)