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

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

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

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

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

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

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

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

Ta có: \(\left(2x-1\right)^2\ge0\)

\(\Rightarrow\) B nhỏ nhất khi \(4x^2-6x+1\)có giá trị nhỏ nhất

Mà: \(4x^2-6x+1=4\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)-\dfrac{5}{4}=4\left(x-\dfrac{3}{4}\right)^2-\dfrac{5}{4}\ge\dfrac{-5}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)

\(\Rightarrow\min\limits_{\left(4x^2-6x+1\right)}=\dfrac{-5}{4}.\)  khi \(x=\dfrac{3}{4}\)

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

\(\Rightarrow\min\limits_B=\dfrac{-5}{4}:\dfrac{1}{4}=\dfrac{-5}{4}.4=-5\)  Khi \(x=\dfrac{3}{4}\)

Ta có: 

 B nhỏ nhất khi có giá trị nhỏ nhất

Mà: ⇔x=34⇔x=34

x=34x=34

⇒minB=−54:14=−54.4=−5⇒minB=−54:14=−54.4=−5  Khi 

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

Biểu thức nào em?

\(\text{a) }\dfrac{x^2+x+1}{x^2+2x+1}\\ =\dfrac{x^2+2x-x+1+1-1}{x^2+2x+1}\\ =\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\\ =1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\left(1\right)\\ Đặt\text{ }\dfrac{1}{x+1}=y\\ \Rightarrow\left(1\right)=1-y+y^2\\ =y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(y-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{2}=0\\ \Leftrightarrow y=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{ 1}{x+1}=\dfrac{1}{2}\\ \Leftrightarrow x+1=2\\ \Leftrightarrow x=1\\ Vậy\text{ }GTNN\text{ }của\text{ }phân\text{ }thức\text{ }là\text{ }\dfrac{3}{4}\text{ }khi\text{ }x=1\)

\(\text{b) }\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\\ =\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\left(1\right)\\ Đặt\text{ }-\dfrac{1}{2x-1}=y\\ \Rightarrow\left(1\right)=1+y+y^2\\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(y+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y+\dfrac{1}{2}=0\\ \Leftrightarrow y=-\dfrac{1}{2}\\ \Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\text{ }GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là\text{ }\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{3}{2}\)

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)