Tham khảo:

CHÚC EM HỌC TỐT NHÁ

hi sư huynh đệ mới lớp 6

\(A=\frac{x^2-2x+2007}{2007x^2},\left(x\ne0\right)\)

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

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

Chúc bạn học tốt !!!

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12007" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">12007 - 22007x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">22007x + 1x2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">1x2

1x2" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">1x2 - 22007x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">22007x + 120072" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">120072 ) + (12007−120072" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">12007−120072 )

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1x−12007" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">1x−12007 = 0

<=> x = 2007

Vậy x = 2007 thì Amin

bài này từng có trên violimpic đấy bạn

dễ ợt 2008

giải đi chứ

Cho mk sửa lại đề nha!

Tìm giá trị nhỏ nhất của biểu thức:

A = | x - 2008 | + | x - 1|

\(Vì \) \(\left|x-2008\right|+\left|x-1\right|\ge0\forall x\)

\(\Rightarrow\left|x-2008\right|+\left|x-1\right|\ge\left|x-2008+x-1\right|=2007\)

\(\Rightarrow\)\(A \) \(đạt\) \(GTNN=2007\Leftrightarrow\)

Đặt \(A=\left|x-2008\right|+\left|x-2007\right|\)

\(\Rightarrow A=\left|x-2008\right|+\left|2007-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(A=\left|x-2008\right|+\left|2007-x\right|\ge\left|x-2008+2007-x\right|\)

\(\Rightarrow A\ge\left|-1\right|\)

\(\Rightarrow A\ge1.\)

Dấu '' = '' xảy ra khi:

\(\left(x-2008\right).\left(2007-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2008\ge0\\2007-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2008\le0\\2007-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2008\\x\le2007\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2008\\x\ge2007\end{matrix}\right.\end{matrix}\right.\Rightarrow2007\le x\le\)

\(\Rightarrow2007\le x\le2008.\)

Vậy \(MIN_A=1\) khi \(2007\le x\le2008.\)

Chúc bạn học tốt!

Ta có: \(\left(x-1\right)^2\ge0\)

\(\Rightarrow A=\left(x-1\right)^2+2008\ge2008\)

Dấu "=" xảy ra <=> (x-1)^2 = 0 <=> x = 1

Vậy GTNN của A = 2008 tại x = 1