Question

Tìm GTLN và GTNN của :

a ) \(M=a^3+b^3+ab\) biết \(a+b=1\)

b ) \(\left(x^2+x\right)\left(x^2+x-4\right)\)

Answer 1

a ) \(M=a^3+b^3+ab\) biết \(a+b=1\)

\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+ab\)

\(M=a^2-ab+b^2+ab\)

\(M=a^2+b^2\)

Ta có : \(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2+b^2\ge2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge a^2+2ab+b^2=\left(a+b\right)^2=1\)

\(\Rightarrow a^2+b^2\ge\frac{1}{2}\)

Vậy \(Min_M=\frac{1}{2}\Leftrightarrow a=b=\frac{1}{2}\).

b ) \(N=\left(x^2+x\right)\left(x^2+x-4\right)=\left[\left(x^2+x-2\right)+2\right]\left[\left(x^2+x-2\right)-2\right]=\left(x^2+x-2\right)^2-4\ge-4\)

Vậy \(Min_N=-4\)\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-2\end{array}\right.\).