Đặt \(\left|3x-1\right|=a\)

\(\Rightarrow\left(3x-1\right)^2=a^2\)

\(\Rightarrow A=a^2-4a+5\)

Biến đổi \(A\)ta được \(A=a^2-4a+4+1=\left(a-2\right)^2+1\ge1\)

Dấu " = " xảy ra \(\Leftrightarrow a-2=0\Leftrightarrow\left|3x-1\right|=2\Leftrightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}}\)

Vậy GTNN của \(A=1\)tại \(\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

\(\text{Đặt }\left|3x-1\right|=a\)

\(\Rightarrow\left(3x-1\right)^2=a^2\)

\(\Rightarrow a=a^2-4a+5\)

\(\text{Biến đổi A ta được }a=\left(a-2\right)^2+1\ge1\)

\(\text{Dấu "=" xảy ra khi }a-2=0=\left|3x-1\right|=2\Rightarrow\hept{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

\(\text{Vậy min A=1}\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)

dk 3x+2 

P= \(\frac{x\left(3x-1\right)}{3x+2}.\frac{3x+2}{\left(3x-1\right)x^2+4\left(3x-1\right)}=\frac{x\left(3x-1\right)}{3x+2}.\frac{3x+2}{\left(3x-1\right)\left(x^2+4\right)}=\)\(\frac{x}{x^2+4}\)

dk \(\hept{\begin{cases}3x-1\ne0\\3x+2\ne0\end{cases}< =>\hept{\begin{cases}x\ne\frac{1}{3}\\x\ne\frac{-2}{3}\end{cases}}}\)(1)

P(x²+4) = x <=> Px²-x+4P=0

để phương trình trên có nghiệm thỏa mãn (1) <=> \(\hept{\begin{cases}P\frac{1}{3^2}-\frac{1}{3}+4P\ne0\\P\frac{4}{9}+\frac{2}{3}+4P\ne0\\1^2-4.P.\left(4P\right)\ge0\end{cases}< =>\hept{\begin{cases}P\ne\frac{3}{37}\\P\ne\frac{-3}{20}\\\frac{-1}{4}\le P\le\frac{1}{4}\end{cases}}}\)

Vậy P max = 1/4 khi \(\frac{1}{4}x^2-x+1=0< =>x=2\)

P min = -1/4 khi \(\frac{-1}{4}x^2-x-1=0< =>x=-2\)

đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)

\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)

đẳng thức khi y=-6 thủa mãn đk (*)

Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)

âm vô hạn

la bao nhiêu

\(A=\frac{x^2-3x+4}{\left(x-1\right)^2}=\frac{x^2+x-4x+4}{\left(x-1\right)^2}=\frac{x\left(x+1\right)+4\left(x+1\right)}{\left(x+1\right)^2}=\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)^2}=\frac{x+4}{x+1}\)

ĐKXĐ: x khác 1

\(A=\frac{x^2-3x+4}{x^2-2x+1}=\frac{x^2-2x+1-x+1+2}{x^2-2x+1}=1+\frac{-\left(x-1\right)}{\left(x-1\right)^2}+\frac{2}{\left(x-1\right)^2}\)

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

đặt \(m=\frac{1}{x-1}\Rightarrow A=1+-m+2m^2=2.\left(m^2-\frac{m.1}{2}+\frac{1}{16}\right)+\frac{7}{8}\)

\(A=2.\left(m-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\)

dấu = xảy ra khi \(m-\frac{1}{4}=0\)

\(\Rightarrow m=\frac{1}{4}=\frac{1}{x-1}\Rightarrow x=5\)

p/s: ko chắc lắm, 60% thôi >:

lỗi:

dòng thứ 4: \(1+\left(-m\right)\)quên dấu ngoặc =.=

thiếu kết luận. Vậy MinA\(=\frac{7}{8}\Leftrightarrow x=5\)

sorry :(

-Đề thiếu?

MinC = 3/4 (khi x = -3/2)

\(C=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4+3x^2-3x^3-9x+3x^2+9}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ C=\dfrac{\left(x^2+3\right)^2-9x^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}\\ C=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

Với mọi x t có :

\(\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|\ge0\)

\(\Leftrightarrow2\left|3x-1\right|-4\ge-4\)

\(\Leftrightarrow A\ge-4\)

Dấu "=" xảy ra khi : \(\left|3x-1\right|=0\)

\(\Leftrightarrow x=\frac{1}{3}\)

Vậy \(A_{Min}=-4\Leftrightarrow x=\frac{1}{3}\)

\(2x^2-3x+1\)

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

\(=2\left[\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{16}+\frac{1}{2}\right]\)

\(=2\left[\left(x-\frac{3}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left(x-\frac{3}{4}\right)^2-\frac{1}{8}\ge\frac{1}{8}\)

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

Ta có \(2x^2\ge0\)giả sử x=0

           \(-3x\ge0\)giả sử x cũng= 0

          1>0

\(\Rightarrow2x^2-3x+1\ge1\)

vậy gtnn của:\(2x^2-3x+1\)là 1