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22 tháng 1 2024

Đặt \(x-1=t\Rightarrow x=t+1\)

\(A=\dfrac{2\left(t+1\right)^2-6\left(t+1\right)+5}{t^2}=\dfrac{2t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+2=\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

\(A_{min}=1\) khi \(t=1\Rightarrow x=2\)

17 tháng 3 2018

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)

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

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

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

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

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

vì (x-1)2 ≥ 0 ∀ x

⇔ (x-1)2 +4 ≥ 4

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

\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)

⇔ A \(\le\dfrac{7}{2}\)

⇔ Min A =\(\dfrac{7}{2}\)

khi x-1=0

⇔ x=1

vậy ....

17 tháng 3 2018

Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)

\(B=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)

\(B=2-\dfrac{3}{x^2-8x+16+6}\)

\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)

\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)

18 tháng 3 2018

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

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 \(x=1\)

18 tháng 3 2018

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

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 \(x=\dfrac{3}{2}\)

20 tháng 11 2017

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

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

7 tháng 3 2018

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

7 tháng 3 2018

ủa thế còn con B....

8 tháng 3 2018

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 = \(\dfrac{-3}{4}\)

12 tháng 5 2018

bt2.

A=[2(4x^2+4x+5)-2]/(4x^2+4x+5)

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

A>=2-2/4=3/2

khi x=-1/4

12 tháng 6 2018

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

*Min A:

Ta có: \(A=\dfrac{2x+1}{x^2+2}\)

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

\(=\dfrac{\left(x+2\right)^2}{2\left(x^2+1\right)}+\dfrac{1}{2}\ge\dfrac{1}{2},\forall x\in R\)

Vậy \(Min_A=\dfrac{1}{2}khi\left(x+2\right)^2=0\)

\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

*Max A:

Ta có: \(A=\dfrac{2x+1}{x^2+2}\)

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

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

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

\(=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le0,\forall x\in R\)

Vậy \(Max_A=1khi\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu '=' xảy ra khi x=3/2

c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)

=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)

=>C>=-12/7

Dấu '=' xảy ra khi x=1/2