Đề sai, biểu thức này chỉ tồn tại max, ko tồn tại min

A= \(\dfrac{x^2-4x+1}{x^2}\)

ĐKXĐ x≠0

A= \(\dfrac{x^2}{x^2}-\dfrac{4x}{x^2}+\dfrac{1}{x^2}\)

=\(1-\dfrac{4}{x}+\dfrac{1}{x^2}\)

đặt \(\dfrac{1}{x}=y\) ta có

1-4y+y²

= y²-4y+1

=(y²-4y+4)-3

= (y-2)² -3

do (y-2)² ≥ 0 ∀x

=> (y-2)² -3 ≥ -3

=> A ≥ -3

=> Amin =-3dấu '=' xảy ra khi

y-2=0

=> y=2

=> \(\dfrac{1}{x}=2\)

=> x=\(\dfrac{1}{2}\)

vậy GTNN A =-3 khi x=\(\dfrac{1}{2}\)

b) ĐKXĐ x ≠\(\dfrac{1}{2}\)

B = \(\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\)

=\(\dfrac{4x^2-6x+1-1+1}{\left(2x-1\right)^2}\)

= \(\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\)

=\(\dfrac{\left(2x-1\right)^2-\left(2x-1\right)-1}{\left(2x-1\right)^2}\)

= \(\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\)

= \(1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\)

đặt \(-\dfrac{1}{2x-1}=y\) ta có

1+y+y²

= \(y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\)

=\(\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

do \(\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\)

=> \(\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=> B ≥\(\dfrac{3}{4}\)

GTNN B =\(\dfrac{3}{4}\)dấu '=' xảy ra khi

y=-\(\dfrac{1}{2}\)

⇔\(-\dfrac{1}{2x-1}=-\dfrac{1}{2}\)

⇔2x-1=2

⇔2x=3

⇔x=\(\dfrac{3}{2}\) (tm)

vậy GTNN B=\(\dfrac{3}{4}\) khi x= \(\dfrac{3}{2}\)

a.

\(A=\dfrac{x^2-4x+1}{x^2}\)

\(\Rightarrow A=\dfrac{x^2-4x+4-3}{x^2}\)

\(\Rightarrow A=\dfrac{\left(x-2\right)^2-3}{x^2}\)

Ta có: \(\left(x-2\right)^2-3\ge-3\)

\(\Rightarrow x=2\)

Khi đó ta được Min A = \(\dfrac{\left(2-2\right)-3}{2^2}\ge\dfrac{-3}{4}\)

Vậy Min A = \(\dfrac{-3}{4}\)

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

\(=\dfrac{x\left(x^2+x+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)

P/s : Mik nghĩ là \(\left(2x+1\right)^2\)

\(C=x+\dfrac{1}{4x}+\dfrac{x}{\left(2x+1\right)^2}=\left[\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\right]+\dfrac{3}{4}\left(x+\dfrac{1}{4x}\right)-\dfrac{1}{8}\)

AD BĐT AM - GM ta được : \(\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\ge4\sqrt[4]{\dfrac{1}{16^3}}=\dfrac{1}{2}\)

\(x+\dfrac{1}{4x}\ge2\sqrt{\dfrac{1}{4}}=1\) 

Suy ra : \(C\ge\dfrac{1}{2}+\dfrac{3}{4}.1-\dfrac{1}{8}=\dfrac{9}{8}\)

" = " \(\Leftrightarrow x=\dfrac{1}{2}\)

\(C=\dfrac{5}{3-\left(4x+1\right)^2}\)

Điều kiện xác định khi 

\(3-\left(4x+1\right)^2\ne0\Leftrightarrow\left[{}\begin{matrix}4x+1\ne\sqrt[]{3}\\4x+1\ne-\sqrt[]{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\sqrt[]{3}-1}{4}\\x\ne\dfrac{-\sqrt[]{3}-1}{4}\end{matrix}\right.\)

Ta có :

\(\left(4x+1\right)^2\ge0,\forall x\)

\(\Leftrightarrow3-\left(4x+1\right)^2\le3\)

\(\Leftrightarrow C=\dfrac{5}{3-\left(4x+1\right)^2}\ge\dfrac{5}{3}\)

Vậy \(GTNN\left(C\right)=\dfrac{5}{3}\left(tạix=-\dfrac{1}{4}\right)\)

\(B=\left(2x\right)^2+2\left(y-1\right)^2-5\)

vì \(\left\{{}\begin{matrix}\left(2x\right)^2\ge0,\forall x\\2\left(y-1\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\Rightarrow B=\left(2x\right)^2+2\left(y-1\right)^2-5\ge-5\)

Dấu "=" xảy tại khi

\(\left\{{}\begin{matrix}2x=0\\2\left(y-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy \(GTNN\left(B\right)=-5\left(tạix=0;y=1\right)\)

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

Do \(\dfrac{\left(x-1\right)^2}{x^2+1}\ge0\forall x\)

\(\Rightarrow B=\dfrac{\left(x-1\right)^2}{x^2+1}+2\ge2\forall x\)

Dấu "=" xảy ra khi :

\(\dfrac{\left(x-1\right)^2}{x^2+1}=0\\ \Leftrightarrow\left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)

Vậy \(B_{\left(Min\right)}=2\) khi \(x=1\)

\(A=\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}=\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\)

Đặt \(-\dfrac{1}{2x-1}=y\)

\(\Rightarrow A=1+y+y^2\\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Do \(\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra khi:

\(\left(y+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y+\dfrac{1}{2}=0\\ \Leftrightarrow y=-\dfrac{1}{2}\\ \Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(A_{\left(Min\right)}=\dfrac{3}{4}\) khi \(x=\dfrac{3}{2}\)