I don't now

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sorry 

......................

Tìm GTNN của A = \(\frac{3x^2-8x+6}{x^2-2x+1}\)

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

\(A=\frac{3x^2-6x+1-2x+5}{x^2-2x+1}\)

\(A=3-\frac{2x-5}{\left(x-1\right)^2}\)

Để A đạt GTNN => \(\frac{2x-5}{\left(x-1\right)^2}\)phải lớn nhất

\(\Rightarrow2x-5\)đạt Max 

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

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

Đặt \(x-1=y\left(y\ne0\right)\)

\(\Rightarrow x=y+1\)

\(A=\frac{3\left(y+1\right)^2-8\left(y+1\right)+6}{y^2}\)

\(A=\frac{3\left(y^2+2y+1\right)-8y-8+6}{y^2}\)

\(A=\frac{3y^2+6y+3-8y-8+6}{y^2}\)

\(A=\frac{3y^2-2y+1}{y^2}\)

\(A=\frac{3y^2}{y^2}-\frac{2y}{y^2}+\frac{1}{y^2}\)

\(A=3-\frac{2}{y}+\frac{1}{y^2}\)

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

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

Mà \(\left(\frac{1}{y}-1\right)^2\ge0\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi :

\(\frac{1}{y}-1=0\)

\(\Leftrightarrow\frac{1}{y}=1\)

\(\Leftrightarrow y=1\)

Mà : \(x=y+1\Rightarrow x=2\)

Vậy \(A_{Min}=2\Leftrightarrow x=2\)

A\(\frac{2x^2-4x+2+x^2-4x+4}{x^2-2x+1}=2+\left(\frac{x-2}{x-1}\right)^2\ge2\)

dấu = xảy ra x=2

chúc ban hk tốt

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

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

Dấu ''='' xảy ra khi GTNN của A=2

Ta có \(\frac{3x^2+8x+6}{x^2+2x+1}=\frac{3x^2+6x+3+2x+2+1}{\left(x+1\right)^2}=\frac{3\left(x+1\right)^2+2\left(x+1\right)+1}{\left(x+1\right)^2}\)

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

Đặt \(\frac{1}{x+1}=t\), biểu thức trở thành: \(t^2+2t+3=\left(t+1\right)^2+2\ge2\)

Vậy GTNN của phân thức là 2, khi t = -1 tức là x = -2.

Ta có:

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

\(\Leftrightarrow A\left(x^2-2x+1\right)=3x^2-8x+6\)

\(\Leftrightarrow\left(3-A\right)x^2+\left(2A-8\right)x+6-A=0\)

Đê pt theo nghiệm x có nghiệm thì

\(\Delta'=\left(A-4\right)^2-\left(3-A\right)\left(6-A\right)\ge0\)

\(\Leftrightarrow A-2\ge0\)

\(\Leftrightarrow A\ge2\)

Vậy GTNN là 2 khi x = 2

x=2

lời giải mk đang làm

bn giải cách lớp 8 đi

Đặt \(x-1=y\) thì \(x=y+1\).Ta có :

\(A=\frac{3\left(y+1\right)^2-8\left(y+1\right)+6}{y^2}=\frac{3y^2-2y+1}{y^2}=3-\frac{2}{y}+\frac{1}{y^2}\)

Lại đặt \(\frac{1}{y}=z\) thì : \(A=3-2z+z^2=\left(z-1\right)^2+2\ge2.\)

Dấu " = " xảy ra khi và chỉ khi \(\left\{\begin{matrix}x-1=1\Leftrightarrow x=2\\y=1\\z=1\end{matrix}\right.\)

Vậy \(Min_A=2\Leftrightarrow\left\{\begin{matrix}x=2\\y=1\\z=1\end{matrix}\right..\)

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

\(S_{min}=-2\) khi \(x=-2\)

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

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

=> MIN S = 2

Dấu "=" xảy ra <=> x - 2 = 0

<=> x = 2

Vậy Min S = 2 khi x = 2

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 \("="\Leftrightarrow x=2\)