\(S=\dfrac{2m^2+7m+23}{m^2+2m+10}\Rightarrow Sm^2+2Sm+10S=2m^2+7m+23\)

\(\Leftrightarrow\left(S-2\right)m^2+\left(2S-7\right)m+10S-23=0\)

\(\Delta=\left(2S-7\right)^2-4\left(S-2\right)\left(10S-23\right)\ge0\)

\(\Leftrightarrow4S^2-16S+15\le0\)

\(\Rightarrow\dfrac{3}{2}\le S\le\dfrac{5}{2}\)

\(S_{min}=\dfrac{3}{2}\) khi \(m=-4\)

\(S_{max}=\dfrac{5}{2}\) khi \(m=2\)

Ta có: \(A=\frac{2m^2-4m+5}{m^2-2m+2}\)

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

\(=\frac{2\left(m-1\right)^2+3}{\left(m-1\right)^2+1}\ge\frac{3}{1}=3\) (do \(\left(m-1\right)^2\ge0\))

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

Vậy \(A_{min}=3\Leftrightarrow m=1\)

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

\(\left(m-1\right)^2+1\ge1\Leftrightarrow\frac{1}{\left(m-1\right)^2+1}\le1\)

\(\Rightarrow A\le3\)

 \("="\Leftrightarrow m=1\)

chết làm lộn r-_-

đặt 2m/(m^2+1)=a. nhân chéo lên rùi đưa về dạng pt bậc hai xét denta lớn hơn bằng 0.=>min,mã. OK!

Bài 1:

\(c,\text{PT có 2 }n_0\text{ phân biệt }\Leftrightarrow\Delta'=2^2-2m>0\Leftrightarrow2m< 4\Leftrightarrow m< 2\)

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