a) Vì \(\left(2x+\frac{1}{4}\right)^4\ge0\forall x\)

\(\Rightarrow A\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{8}\)

b) \(B=-\left(\frac{4}{9}x-\frac{2}{15}\right)^6+3\)

\(B=3-\left(\frac{4}{9}x-\frac{2}{15}\right)^6\)

Vì \(\left(\frac{4}{9}x-\frac{2}{15}\right)^6\ge0\forall x\)

\(\Rightarrow B\le3\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{4}{9}x-\frac{2}{15}=0\Leftrightarrow x=\frac{3}{10}\)

với mọi x thì (2x+1/4)⁴>=0 (lớn  hơn hoặc bằng )

A=(2x+1/4)⁴-1>=-1

để A đạt GTNN thì (2x+1/4)⁴=0

2x+1/4=0 =>x=-1/8

\(A=\left(x+3\right)\left(x-4\right)+7=x^2-x-5=\left(x^2-x+\frac{1}{4}\right)-\frac{1}{4}-5\)

\(=\left(x-\frac{1}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

"=" <=> x = 1/2

\(B=3-\left(x-1\right)\left(x-2\right)=3-\left(x^2-3x+2\right)\)

\(=3-\left(x-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+2\right)\)

\(=3+\frac{1}{4}-\left(x-\frac{3}{2}\right)^2\le\frac{13}{4}\)

Xảy ra khi x = 3/2

em xét dấu trị tuyệt đối với mũ 2 nhé

a: Ta có: \(-x^2+4x+5\)

\(=-\left(x^2-4x-5\right)\)

\(=-\left(x^2-4x+4-9\right)\)

\(=-\left(x-2\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=2

b: Ta có: \(-x^2-7x+4\)

\(=-\left(x^2+7x-4\right)\)

\(=-\left(x^2+2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{65}{4}\right)\)

\(=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{65}{4}\le\dfrac{65}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{7}{2}\)

a: \(A\ge-5\forall x,y\)

Dấu '=' xảy ra khi x=2 và y=-1