Question

giúp mik bài 5 ạ cảm ơn

Answer 1

a) \(A=x^2+5x+7=\left(x^2+5x+\dfrac{25}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minA=\dfrac{3}{4}\Leftrightarrow x=-\dfrac{5}{2}\)

b) \(B=6x-x^2-5=-\left(x^2-6x+9\right)+4=-\left(x-3\right)^2+4\le4\)

\(maxB=4\Leftrightarrow x=3\)

c) \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)

\(minC=-36\Leftrightarrow x^2+5x=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)