Để A = 5012-2013:(2014-x)có GTNN (giá trị nhỏ nhất)

thì 2013:(2014-x) có TGLN (giá trị lớn nhất)

tương đương 2014 - x có GTNN

Vì 2014 - x là số chia nên 2014 - x khác 0.

Do đó 2014 -x có GTNN là 1 => x = 2014 - 1 = 2013

Vậy x là 2013

Sửa đề: \(A=\left(2x-1\right)\left(2x^2-3x-1\right)\left(x-1\right)+2005\)

\(=\left(2x^2-3x+2\right)\left(2x^2-3x-1\right)+2005\)

\(=\left(2x^2-3x\right)^2+\left(2x^2-3x\right)+2+2005\)

\(=\left(2x^2-3x\right)^2+\left(2x^2-3x\right)+\frac14+2006,75=\left(2x^2-3x+\frac12\right)^2+2006,75\ge2006,75\forall x\)

Dấu '=' xảy ra khi \(2x^2-3x+\frac12=0\)

=>\(x^2-\frac32x+\frac14=0\)

=>\(x^2-2\cdot x\cdot\frac34+\frac{9}{16}-\frac{7}{16}=0\)

=>\(\left(x-\frac34\right)^2=\frac{7}{16}\)

=>\(\left[\begin{array}{l}x-\frac34=\frac{\sqrt7}{4}\\ x-\frac34=-\frac{\sqrt7}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt7+3}{4}\\ x=\frac{-\sqrt7+3}{4}\end{array}\right.\)

a)-19

b)22

lập bảng xét dấu đi

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

a: \(A=x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

b: \(B=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x-10=0

=>x=10

A =|3x-4| + |5x-7| -x +2025

- Nếu x < \(\dfrac{4}{3}\):

\(\Rightarrow\) \(\left\{{}\begin{matrix}3x-4< 0\\5x-7< 0\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}\text{|}3x-4\text{|}=-3+4\\\text{|}5x-7\text{|}=-5x+7\end{matrix}\right.\) 

\(\Rightarrow\) \(A=-3x+4-5x+7-x+2025\) 

Vì x \(< \dfrac{4}{3}\) \(\Rightarrow\) \(9x< 12\) \(\Rightarrow\) \(-9x>-12\) 

\(\Rightarrow\) \(-9x+2036>2024\) 

\(\Rightarrow\) A \(>2024\) ( Loại)

Nếu \(\dfrac{4}{3}\) \(\le\) x \(< \dfrac{7}{5}\) 

\(\Rightarrow\) \(\left\{{}\begin{matrix}3x-4>0\\5x-7< 0\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}\text{|}3x-4\text{|}=3x-4\\\text{|}5x-7\text{|}=-5x+7\end{matrix}\right.\) 

\(\Rightarrow\) A= \(-3x-4-5x+7-x+2025\) 

       =   \(-3x+2028\) 

Ta có: \(\dfrac{4}{3}\) \(\le x\) \(\Rightarrow\) \(-3x\) \(>\dfrac{-21}{5}\) 

\(\Rightarrow\) 2024 \(\ge\) \(-3x+2028>\dfrac{10119}{5}\) ( loại)

Nếu x :

\(\ge\dfrac{7}{5}\\ \Rightarrow\left\{{}\begin{matrix}3x-4>0\\5x-7>0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\text{|}3x-4\text{|}=3x-4\\\text{|}5x-7\text{|}=5x-7\end{matrix}\right.\\ \Rightarrow A=3x-4+5x-7-x+2025\) 

  \(=7x+2014\) 

Vì \(x\ge\dfrac{7}{5}\) \(\Rightarrow\) \(7x\ge\dfrac{49}{5}\) 

\(\Rightarrow\) \(7x+2014\) \(\ge\dfrac{19}{5}+2014=\dfrac{10119}{5}\) 

\(\Rightarrow\) A \(\ge\) \(\dfrac{10119}{5}\) (  t/m)

Vậy A đạt GTNN khi A bằng \(\dfrac{10119}{5}\)

Dấu "=" xảy ra khi  \(x=\dfrac{7}{5}\)

\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)