bài này ta có thể giải theo 2 cách 

ta có A = \(\frac{x^2-2x+2011}{x^2}\)

= \(\frac{x^2}{x^2}\)- \(\frac{2x}{x^2}\)+ \(\frac{2011}{x^2}\)

= 1 - \(\frac{2}{x}\)+ \(\frac{2011}{x^2}\)

đặt \(\frac{1}{x}\)= y ta có 

A= 1- 2y + 2011y^2 

cách 1 : 

A = 2011y^2 - 2y + 1 

= 2011 ( y^2 - \(\frac{2}{2011}y\)+ \(\frac{1}{2011}\)) 

= 2011( y^2 - 2.y.\(\frac{1}{2011}\)+ \(\frac{1}{2011^2}\)- \(\frac{1}{2011^2}\) + \(\frac{1}{2011}\)) 

= 2011 \(\left(\left(y-\frac{1}{2011}\right)^2\right)+\frac{2010}{2011^2}\)

= 2011\(\left(y-\frac{1}{2011}\right)^2\)+ \(\frac{2010}{2011}\)

vì ( y - \(\frac{1}{2011}\)) ²>=0 

=> 2011\(\left(y-\frac{1}{2011}\right)^2\)+ \(\frac{2010}{2011}\)> = \(\frac{2010}{2011}\)

hay A >=\(\frac{2010}{2011}\)

cách 2  

A = 2011y^2 - 2y + 1 

= ( \(\sqrt{2011y^2}\)) - 2 . \(\sqrt{2011y}\). \(\frac{1}{\sqrt{2011}}\)+ \(\frac{1}{2011}\)+ \(\frac{2010}{2011}\)

= \(\left(\sqrt{2011y}-\frac{1}{\sqrt{2011}}\right)^2\)+ \(\frac{2010}{2011}\)

vì \(\left(\sqrt{2011y}-\frac{1}{\sqrt{2011}}\right)^2\)> =0 

nên \(\left(\sqrt{2011y}-\frac{1}{\sqrt{2011}}\right)^2\)+ \(\frac{2010}{2011}\)>= \(\frac{2010}{2011}\)

hay A >= \(\frac{2010}{2011}\)

ta thay M=(2011-x-1)/(2011-x)                  =1-1/(2011-x)                                         de M nho nhat thi 1/(2011-x) lon nhat suyra 2011-x nho nhat   va nguyen duong suy ra x=2010      suy ra gia tri nho nhat cua M=0                  

k biet lam

\(\text{Ta có:}x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5\ge0+5=5\)

\(P=\frac{1}{x^2+2x+6}\ge\frac{1}{5}\Rightarrow\text{GTLN của }P\text{ là:}\frac{1}{5}\text{ khi: }x=\frac{1}{5}\)

C1 : 

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=1\)

C2 : 

\(B=\frac{x^2+x+1}{x^2+2x+1}\)\(\Leftrightarrow\)\(Bx^2-x^2+2Bx-x+B-1=0\)

\(\Leftrightarrow\)\(\left(B-1\right)x^2+\left(2B-1\right)x+\left(B-1\right)=0\)

+) Nếu \(B=1\) thì \(x=0\)

+) Nếu \(B\ne1\) thì pt có nghiệm \(\Leftrightarrow\)\(\Delta\ge0\)

                                                        \(\Leftrightarrow\)\(\left(2B-1\right)^2-4\left(B-1\right)\left(B-1\right)\ge0\)

                                                        \(\Leftrightarrow\)\(4B^2-4B+1-4B^2+8B-4\ge0\)

                                                        \(\Leftrightarrow\)\(4B-3\ge0\)

                                                        \(\Leftrightarrow\)\(B\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=1\)

Ta có :

 \(B=\frac{x^2-2x+2011}{x^2}\)

\(B=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2011}{x^2}\)

\(B=1-\frac{2}{x}+\frac{2011}{x^2}\)

\(B=\left(\frac{\sqrt{2011}^2}{x^2}-\frac{2}{x}+\frac{1}{2011}\right)+\frac{2010}{2011}\)

\(B=\left(\frac{\sqrt{2011}}{x}-\frac{1}{\sqrt{2011}}\right)^2+\frac{2010}{2011}\)

Mà : \(\left(\frac{\sqrt{2011}}{x}-\frac{1}{\sqrt{2011}}\right)^2\ge0\forall x\)

\(\Rightarrow B\ge\frac{2010}{2011}\)

Dấu "=" xảy ra khi :

\(\frac{\sqrt{2011}}{x}-\frac{1}{\sqrt{2011}}=0\)

\(\Leftrightarrow x=2\sqrt{2011}\)

Vậy \(MinB=\frac{2010}{2011}\Leftrightarrow x=2\sqrt{2011}\)

a) Ta có \(x^2+2x+6=\left(x+1\right)^2+5\ge5\)

\(\Rightarrow P\le\frac{1}{5}\)

Dấu "=" xảy ra khi x=-1

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

Đặt \(a=\frac{1}{x+1}\)

\(\Rightarrow Q=1-a+a^2=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(a=\frac{1}{2}\Rightarrow x=1\)

\(P=\frac{1}{x^2+2x+6}\)

để pmin thì \(x^2+2x+6max\)

\(\frac{1}{x^2+2x+6}=\frac{1}{\left(x+1\right)^2+5}\)lớn hơn hoặc bằng 1/5 

=>Pmin=1/5 khi và chỉ khi x=-1