4-3=2( dân chơi mới hiểu)

Chắc là viết thiếu số "1" đấy, sợ lớp 11 còn chưa làm được cơ

\(S=\dfrac{3x^2+8x+6}{x^2+2x+1}=\dfrac{-2\left(x^2+2x+1\right)+x^2+4x+4}{x^2+2x+1}=-2+\left(\dfrac{x+2}{x+1}\right)^2\ge-2\)

\(S_{min}=-2\) khi \(x=-2\)

sao lại ra -2 thé thầy ??

\(S=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{2x^2-4x+2+x^2-4x+4}{x^2-2x+1}\)

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

=> MIN S = 2

Dấu "=" xảy ra <=> x - 2 = 0

<=> x = 2

Vậy Min S = 2 khi x = 2

\(\frac{3x^2-8x+6}{x^2-2x+1}\)

=\(\frac{2x^2-x^2-4x-4x+2+4}{x^2-2x+1}\)

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

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

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

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

Vì \(\frac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge0\)  với mọi x

<=>\(2+\frac{\left(x-2\right)^2}{\left(x-1\right)^2}\) > 2 với mọi x

Dấu "=" xảy ra khi và chỉ khi x=-2 thì Min =2

Vậy Min=2

Tử \(x^4+2x^3+8x+16\)

\(=x^4-2x^3+4x^2+4x^3-8x^2+16x+4x^2-8x+16\)

\(=x^2\left(x^2-2x+4\right)+4x\left(x^2-2x+4\right)+4\left(x^2-2x+4\right)\)

\(=\left(x^2+4x+4\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)^2\left(x^2-2x+4\right)\)

Mẫu \(x^4-2x^3+8x^2-8x+16\)

\(=x^4-2x^3+4x^2+4x^2-8x+16\)

\(=x^2\left(x^2-2x+4\right)+4\left(x^2-2x+4\right)\)

\(=\left(x^2+4\right)\left(x^2-2x+4\right)\)

Thay tử và mẫu vào ta có:\(\frac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(x^2+4\right)\left(x^2-2x+4\right)}=\frac{\left(x+2\right)^2}{x^2+4}\ge0\)

Dấu "=" khi \(\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Vậy Min=0 khi x=-2

Ta có: \(\frac{3x^2-8x+6}{x^2-2x+1}=B\)

=>\(B=\frac{3x^2-6x+3-2x+3}{x^2-2x+1}=3+\frac{-2x+3}{\left(x-1\right)^2}\)

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

\(=\left(\frac{1}{x-1}-1\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi \(\frac{1}{x-1}-1=0\)

=>x-1=1

=>x=2

\(B=2x^2+y^2-8x+2xy-4y+2025\)

\(=x^2+2xy+y^2-4x-4y+x^2-4x+2025\)

\(=\left(x+y\right)^2-4\left(x+y\right)+4+x^2-4x+4+2017\)

\(=\left(x+y-2\right)^2+\left(x-2\right)^2+2017\ge2017\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-2=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=-x+2=-2+2=0\end{cases}\)

b: \(B=\sqrt{x^2-8x+18}-1\)

\(=\sqrt{\left(x-4\right)^2+2}-1\)

(x-4)^2+2>=2

=>\(\sqrt{\left(x-4\right)^2+2}>=\sqrt{2}\)

=>B>=căn 2-1

Dấu = xảy ra khi x=4

a: \(D=3+\sqrt{2x^2-8x+33}\)

\(=3+\sqrt{2\left(x^2-4x+\dfrac{33}{2}\right)}\)

\(=\sqrt{2\left(x^2-4x+4\right)+25}+3\)

\(=\sqrt{2\left(x-2\right)^2+25}+3>=5+3=8\)

Dấu = xảy ra khi x=2

Cứu