Question

[Nguyễn Hoàng Hà]

Tim GTNN

A=4X²-X-2

B= \(\dfrac{2X^2+6X-3}{5}\)

C= X⁴+4X-1

D= 4X²+\(\dfrac{9}{X^2}\) với x khác 0

Answer 1

[Akai Haruma]

Lời giải:

a)

Ta có: \(A=4x^2-x-2=(2x)^2-2.2x.\frac{1}{4}x+(\frac{1}{4})^2-\frac{33}{16}\)

\(=(2x-\frac{1}{4})^2-\frac{33}{16}\)

Vì \((2x-\frac{1}{4})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A\ge 0-\frac{33}{16}=-\frac{33}{16}\)

Vậy GTNN của $A$ là $\frac{-33}{16}$ khi $x=\frac{1}{8}$

b)

\(B=\frac{2x^2+6x-3}{5}=\frac{2(x^2+3x+\frac{9}{4})-\frac{15}{2}}{5}\)

\(=\frac{2(x+\frac{3}{2})^2-\frac{15}{2}}{5}\geq \frac{2.0-\frac{15}{2}}{5}=\frac{-3}{2}\)

Vậy \(B_{\min}=\frac{-3}{2}\Leftrightarrow (x+\frac{3}{2})^2=0\Leftrightarrow x=\frac{-3}{2}\)

Answer 2

[Akai Haruma]

c)

\(C=x^4+4x-1\)

\(=x^4-2x^2+1+2x^2+4x-2\)

\(=(x^2-1)^2+2(x^2+2x+1)-4\)

\(=(x^2-1)^2+2(x+1)^2-4\)

\(=(x-1)^2(x+1)^2+2(x+1)^2-4=(x+1)^2[(x-1)^2+2]-4\)

Thấy rằng:

\((x+1)^2\geq 0; (x-1)^2+2>0\Rightarrow (x+1)^2[(x-1)^2+2]\geq 0\)

\(\Rightarrow C\geq 0-4=-4\)

Vậy $C_{\min}=-4$ khi \((x+1)^2=0\Leftrightarrow x=-1\)

d)

\(D=4x^2+\frac{9}{x^2}=(2x)^2+(\frac{3}{x})^2-2.2x.\frac{3}{x}+12\)

\(=(2x-\frac{3}{x})^2+12\geq 0+12=12\)

Vậy $D_{\min}=12$ khi \(2x-\frac{3}{x}=0\Leftrightarrow x=\pm \sqrt{\frac{3}{2}}\)