Question

Tìm GTNN của

A=3x^2-x+4

B=(x-2)(x-5)(x^2-7x-10)

Answer 1

A=3x²-x+4

\(=3\left(x^2-\frac{x}{3}+\frac{4}{3}\right)\)

\(=3\left(x-\frac{1}{6}\right)^2+\frac{47}{12}\ge0+\frac{47}{12}=\frac{47}{12}\)

Dấu = khi \(x=\frac{1}{6}\)

Vậy MinA=\(\frac{47}{12}\Leftrightarrow x=\frac{1}{6}\)

 

 

 

Answer 2

B=(x-2)(x-5)(x²-7x-10)

=(x²-7x+10)(x²-7x-10)

Đặt t=x²-7x+10 đc:

B=t(t-20)=t²-20t

=t²-20t+100-100

=(t-10)²-100

Thay t=x²-7x+10 ta đc: 

\(B=\left(x^2-7x+10-10\right)-100\ge0-100=-100\)

\(\Rightarrow B\ge-100\)

Dấu = khi \(\left[\begin{array}{nghiempt}x=0\\x=7\end{array}\right.\)

Vậy MinB=-100 khi \(\left[\begin{array}{nghiempt}x=0\\x=7\end{array}\right.\)