Question

[Phùng Gia Huy]

Tìm GTLN hoặc GTNN của biểu thức: B=(1-x)(3x+4)

Answer 1

[Yen Nhi]

\(B\left(1-x\right)\left(3x+4\right)\)

\(\rightarrow B=\frac{1}{3}\left(3-3x\right)\left(3x+4\right)\)

\(\rightarrow B\text{⩽ }\frac{1}{3}\left(\frac{3-3x+3x+4}{2}\right)^2\)

\((BTD\)\(AM-GM)\)

\(\rightarrow B\text{⩽ }\frac{1}{3}.\frac{49}{4}\)

\(\rightarrow B\text{⩽ }\frac{49}{12}\)

Dấu '' = '' xảy ra \(\Leftrightarrow3-3x=3x+4\Leftrightarrow-\frac{1}{6}\)

Vậy \(max\)\(B=\frac{49}{12}\Leftrightarrow x=-\frac{1}{6}\)

Answer 2

[ミ★Zero ❄ ( Hoàng Nhật )]

\(B=\left(1-x\right).\left(3x+4\right)\)

Ta có :

\(B=3x+4-3x^2-4x\)

\(B=-3x^2-x+4\)

\(B=-3\left(x^2+\frac{1}{3}x-\frac{4}{3}\right)\)

\(B=-3\left(x^2+2.\frac{1}{6}.x+\frac{1}{36}-\frac{1}{36}-\frac{4}{3}\right)\)

\(B=-3\left[\left(x+\frac{1}{6}\right)^2-\frac{49}{36}\right]\)

Vì \(\left(x+\frac{1}{6}\right)^2\ge0\)

\(\Rightarrow\left(x+\frac{1}{36}\right)^2-\frac{49}{36}\ge-\frac{49}{36}\)

\(\Rightarrow B\le\frac{49}{12}\)

\(\Rightarrow\)GTLN của B là \(\frac{49}{12}\)Khi \(x=-\frac{1}{6}\)