\(f\left(x\right)=\frac{2x^2-2x+3}{x^2-x+2}=\frac{2\left(x^2-x+2\right)-1}{x^2-x+2}=2-\frac{1}{x^2-x+2}=2-\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\)

Ta thấy : \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\forall x\)

\(\Leftrightarrow\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\ge\frac{1}{\frac{7}{4}}=\frac{4}{7}\forall x\)

\(\Rightarrow f\left(x\right)=2-\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{7}{4}}\ge2-\frac{4}{7}=\frac{10}{7}\forall x\) có GTNN là \(\frac{10}{7}\)

Dấu "=" xảy ra <=> \(\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

Vậy \(f\left(x\right)_{min}=\frac{10}{7}\) tại \(x=\frac{1}{2}\)

Sai rồi bạn!

Tớ nói chơi thôi chứ đúng rồi đó.

Ta có : \(A=\frac{x^2+2x+3}{\left(x+2\right)^2}\) . Đặt \(y=x+2\Rightarrow x=y-2\)

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

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

Đặt \(\frac{1}{y}=z\Rightarrow A=3z^2-2z+1=3\left(z-\frac{1}{3}\right)^2+\frac{2}{3}\ge\frac{2}{3}\)

Dấu đẳng thức xảy ra \(\Leftrightarrow z=\frac{1}{3}\Leftrightarrow y=3\Leftrightarrow x=1\)

Vậy Min A = \(\frac{2}{3}\Leftrightarrow x=1\)

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

\(=\frac{\left(x-1\right)^2}{3\left(x+2\right)^2}+\frac{2}{3}\ge\frac{2}{3}\). Dấu đẳng thức xảy ra khi x = 1.

Vậy Min A = 2/3 <=> x = 1

\(A=\frac{x^2+2x+3}{\left(x+2\right)^2}=\frac{3x^2+6x+9}{3\left(x+2\right)^2}=\frac{\left(x^2-2x+1\right)+\left(2x^2+8x+8\right)}{3\left(x+2\right)^2}\)

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

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy GTNN của A là \(\frac{2}{3}\Leftrightarrow x=1\)

Thiếu ĐKXĐ \(x\ne-2\)

Vì \(\hept{\begin{cases}\left(2x+3\right)^2\ge0\\\left|x^2-\frac{9}{4}\right|\ge0\end{cases}}\)=> \(D\ge3\cdot0+2\cdot0+3,5=3,5\)

Dấu = xảy ra khi       \(x=-\frac{3}{2}\)

Ta có: 

\(D=3\left(2x+3\right)^2+2\left|x^2-\frac{9}{4}\right|+3,5\)

Mà: \(\left(2x+3\right)^2\ge0\)          với mọi x 

   \(\left|x^2-\frac{9}{4}\right|\ge0\)   với mọi x

\(\Rightarrow3\left(2x+3\right)^2+2\left|x^2-\frac{9}{4}\right|\ge0\)

\(\Rightarrow3\left(2x+3\right)^2+2\left|x^2-\frac{9}{4}\right|+3,5\ge3,5\)

Dấu "=" xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}2x+3=0\\x^2-\frac{9}{4}=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\pm\frac{3}{2}\end{cases}\Rightarrow}x=\pm\frac{3}{2}}\)

Vậy: GTNN của D bằng 3,5 khi x = \(\pm\)\(\frac{3}{2}\)

Bạn phuong là sai r, hợp của chúng phải là x=-3/2 mới đúng

ĐK: \(\left(x-2\right)\left(x^2+1\right)+2x\left(x-2\right)\ne0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)^2\ne0\Leftrightarrow x\ne-1;2\)

Ta có: \(A=\frac{x^2\left(x-2\right)+4\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+1\right)}=\frac{x^2+4}{\left(x+1\right)^2}=\frac{t^2-2t+5}{t^2}\left(t=x+1\right)\)

\(=\frac{5}{t^2}-\frac{2}{t}+1=5\left(\frac{1}{t}-\frac{1}{5}\right)^2+\frac{4}{5}\ge\frac{4}{5}\)

Đẳng thức xảy ra khi t = 5 hay x=4

Vậy..

Hy vọng bạn học BĐT Cauchy rồi

\(x\ne-1\)

Đặt \(\left(x+1\right)^2=a>0\Rightarrow P=\frac{\left(a+2\right)\left(a+8\right)}{a}=\frac{a^2+10a+16}{a}\)

\(P=a+\frac{16}{a}+10\ge2\sqrt{a.\frac{16}{a}}+10=18\)

\(\Rightarrow P_{min}=18\) khi \(a=4\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2