ta có \(A=\frac{2x+1}{x^2}=\frac{x^2+2x+1-x^2}{x^2}=\frac{\left(x+1\right)^2}{x^2}-1\)

vì \(\frac{\left(x+1\right)^2}{x^2}\ge0\Rightarrow A\ge-1\)

dấu = xảy ra <=> x=-1

\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}=\frac{x^2-2x.2007+2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}\)

\(=\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)

A min =\(\frac{2006}{2007}\)khi \(x-2007=0\)

\(\Leftrightarrow x=2007\)

\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}\)

\(A=\frac{x^2-2x.2007-2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}\)

\(A=\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)

\(\Rightarrow Amin=\frac{2006}{2007}\)khi \(x-2007=0\)

\(\Rightarrow x=2007\)

Bài 1a) 

\(P\left(x\right)=x^{2018}+4x^2+10\)

VÌ \(x^{2018}\ge0\forall x;4x^2\ge0\forall x\)

\(\Rightarrow x^{2018}+4x^2+10\ge10\forall x\)

Hay \(P\left(x\right)\ge10\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Bài 1b)

\(M\left(x\right)=x^2+x+1\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-1}{2}\)

Bài 2a)

\(Q\left(x\right)=-x^4-1\)

Vì \(-x^4\le0\forall x\)

\(\Rightarrow-x^4-1\le-1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

2, TC: \(\frac{5x^2-4x+4}{x^2}=\frac{4x^2+x^2-4x+4}{x^2}\)\(=\frac{4x^2}{x^2}+\frac{\left(x-2\right)^2}{x^2}=4+\frac{\left(x-2\right)^2}{x^2}\)

Ta có \(\frac{\left(x-2\right)^2}{x^2}\ge0\forall x\left(x\ne0\right)\)\(\Rightarrow4+\frac{\left(x-2\right)^2}{x^2}\ge4\)

Vậy GTNN của A là 4 tại \(\frac{\left(x-2^2\right)}{x^2}=0\Rightarrow x=2\)

Bài này mài kiếm đâu ra z mk hềnh như bài này ta lm oy mk

Ta có: \(P=\frac{2016x^2-2x+1}{x^2}=\frac{2015x^2+\left(x^2-2x+1\right)}{x^2}\)

\(=2015+\frac{\left(x-1\right)^2}{x^2}\ge2015\left(\forall x\ne0\right)\)

Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy Min(P) = 2015 khi x = 1

Ta có : \(P=\frac{2016x^2-2x+1}{x^2}\)

\(=\frac{2015x^2+\left(x-1\right)^2}{x^2}\)

\(=2015+\left(\frac{x-1}{x}\right)^2\)

Vì \(\left(\frac{x-1}{x}\right)^2\ge0\forall x\ne0\)

\(\Rightarrow P\ge2015\forall x\ne0\)

Dấu \("="\) xảy ra \(\Leftrightarrow\left(\frac{x-1}{x}\right)^2=0\)

\(\Leftrightarrow\frac{x-1}{x}=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy \(MinP=2015\Leftrightarrow x=1\)

\(P-2015=\dfrac{\left(x-1\right)^2}{x^2}\ge0\) nên \(P\ge2015\), xảy ra dấu bằng khi x = 1.

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

Đặt \(t=\frac{1}{x}\) ta có: \(A=2011t^2-2t+1\)

\(\Leftrightarrow A=2011t^2-2t+\frac{1}{2011}+\frac{2010}{2011}\)

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

\(\Leftrightarrow A=2011\left(t-\frac{1}{2011}\right)^2+\frac{2010}{2011}\ge\frac{2010}{2011}\)

Đẳng thức xảy ra khi \(t=\frac{1}{2011}\Leftrightarrow x=2011\)

Ta có:\(\frac{x^2-2x+2011}{x^2}\ge\frac{2010}{2011}\Rightarrow2011\left(x^2-2x+2011\right)\ge2010x^2\)

\(\Rightarrow2011x^2-2x2011+2011^2\ge2010^2\)

\(\Rightarrow2011x^2-2x2011+2011-2010x^2\ge0\)

\(\Rightarrow x^2-2x2011+2011^2\ge0\)

\(\Rightarrow\left(x-2011\right)^2\ge0\)(đúng)

\(\Rightarrow\)đpcm

Ta có x²>=0với  mọi x mà x khác 0=>x²>0

A nhỏ nhất  <=>x²-2x+2011nhỏ nhất 

<=>(x-1)²+2009 nhỏ nhất =>giá trị nhỏ nhất 2009