Vì \(A=\frac{x^2-2x+2014}{\left(x+1\right)^2}\)

\(\Rightarrow x^2-2x+2014=A\left(x+1\right)^2\)

\(\Leftrightarrow x^2-2x+2014=Ax^2+2Ax+A\)

\(\Leftrightarrow\left(1-A\right)x^2-2\left(A+1\right)x+\left(2014-A\right)=0\)

\(\Delta=4\left(A+1\right)^2-4\left(1-A\right)\left(2014-A\right)\)

\(=8068A-8052\)

Vì A có GTNN nên phương trình có nghiệm

\(\Leftrightarrow8068A-8052\ge0\Leftrightarrow A\ge\frac{2013}{2017}\)

Dấu "=" khi \(x=\frac{2015}{2}\)

A=\(1-\frac{2}{x}+\frac{2014}{x^2}=1-\frac{2.\sqrt{2014}}{x}.\frac{1}{\sqrt{2014}}+\left(\frac{\sqrt{2014}}{x}\right)^2=\left(\frac{\sqrt{2014}}{x}\right)^2-\frac{2}{x}+\frac{1}{2014}+\frac{2013}{2014}=\left(\frac{\sqrt{2014}}{x}-\frac{1}{\sqrt{2014}}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\)

Vậy Min A là 2013/2014 với x=2014

A= (x+ 1) ( 2x- 1)

A = 2x² + x -1 

\(A=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{\sqrt{2}}{4}+\left(\frac{\sqrt{2}}{4}\right)^2-\frac{9}{8}.\)

 \(A=\left(\sqrt{2}x+\frac{\sqrt{2}}{4}\right)^2-\frac{9}{8}\)

Tự giải nốt phần sau EZ rồi 

`a)`

`A=(x+1)(2x-1)`

`=2x^{2}+x-1`

`=2(x^{2}+(1)/(2)x-(1)/(2))`

`=2(x^{2}+(1)/(2)x+(1)/(16)-(9)/(16))`

`=2(x+(1)/(4))^{2}-(9)/(8)>= -9/8` với mọi `x`

Dấu `=` xảy ra khi :

`x+(1)/(4)=0<=>x=-1/4`

Vậy `min=-9/8<=>x=-1/4`

``

`b)`

`(4x+1)(2x-5)`

`=8x^{2}-18x-5`

`=8(x^{2}-(9)/(4)x-(5)/(8))`

`=8(x^{2}-(9)/(4)x+(81)/(64)-(121)/(64))`

`=8(x-(9)/(8))^{2}-(121)/(8)>= -(121)/(8)` với mọi `x`

Dấu `=` xảy ra khi :

`x-(9)/(8)=0<=>x=9/8`

Vậy `min=-121/8<=>x=9/8`

\(A=2x^2+x-1=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(A_{min}=-\dfrac{9}{8}\) khi \(x=-\dfrac{1}{4}\)

\(B=8x^2-18x-5=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(B_{min}=-\dfrac{121}{8}\) khi \(x=\dfrac{9}{8}\)

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{9}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)

b) \(B=\left(4x+1\right)\left(2x-5\right)=8x^2-18x-5=8\left(x^2-\dfrac{9}{4}x+\dfrac{81}{64}\right)-\dfrac{121}{8}=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\)

Vì \(8\left(x-\dfrac{9}{8}\right)^2\ge0\Rightarrow8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)

\(ĐTXR\Leftrightarrow x=\dfrac{9}{8}\)

a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

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

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)

1. a) Ta có:

|x-3| > 0

=> |x-3| + 2 > 2

=> (|x-3| + 2)² > 2² = 4

|y+3| > 0

=> P = (|x-3|+2)² + |y+3| + 2007 > 4 + 0 + 2007 = 2011

=> GTNN của P là 2011

<=> x-3 = y+3 = 0

<=> x = 3; y = -3.

x+2/2013+x+1/2014=x/2015+x-1/2016

a) \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|=2007\)

Ta có: \(\left|x-3\right|\ge0\forall x\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2\ge\left(0+2\right)^2=2^2=4\)

Lại có: \(\left|y+3\right|\ge0\forall y\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2+\left|y+3\right|\ge4+0=4\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2+\left|y+3\right|+2007\ge4+2007=2011\)

 \(\Rightarrow P_{MIN}=2011\)

Dấu "=" xảy ra khi \(\Leftrightarrow\orbr{\begin{cases}\left|x-3\right|=0\\\left|y+3\right|=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\y=-3\end{cases}}}\)

Vậy \(P_{MIN}=2011\) tại \(\orbr{\begin{cases}x=3\\y=-3\end{cases}}\)