Bài 1:

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

A=|m+1|+|m-1|=|m+1|+|1-m|>=|m+1+1-m|=2

Dấu = xảy ra khi -1<=m<=1

B=|2a-1|+|2a-3|=|2a-1|+|3-2a|>=|2a-1+3-2a|=2

Dấu = xảy ra khi 1/2<=a<=3/2

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

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

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

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

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

\(S_{min}=\dfrac{3}{2}\) khi \(m=-4\)

\(S_{max}=\dfrac{5}{2}\) khi \(m=2\)

\(\left\{{}\begin{matrix}x+y=2m-1\left(1\right)\\x^2+y^2=m^2+2m-3\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left(x+y\right)^2-2xy=m^2+2m-3\)

\(\Leftrightarrow\left(2m-1\right)^2-m^2-2m+3=2xy\)

\(\Leftrightarrow2xy=3m^2-6m+4\)

\(P_{min}\Leftrightarrow3m^2-6m+4\left(min\right)\)

\(3\left(m^2-2m+\dfrac{4}{3}\right)=3\left(m^2-2m+1+\dfrac{1}{3}\right)=3\left[\left(m-1\right)^2+\dfrac{1}{3}\right]=3\left(m-1\right)^2+1\ge1\)

\("="\Leftrightarrow m=1\)

\(\Leftrightarrow m.cosx+\left(2m-1\right)sinx+3-m=y\left(cosx+sinx-2\right)\)

\(\Leftrightarrow\left(m-y\right)cosx+\left(2m-y-1\right)sinx=m-2y-3\)

Pt có nghiệm khi:

\(\left(m-y\right)^2+\left(2m-y-1\right)^2\ge\left(m-2y-3\right)^2\)

\(\Leftrightarrow2y^2+\left(2m+10\right)y-4m^2-2m+8\le0\)

\(\Rightarrow\dfrac{-m-5-\sqrt{9m^2+14m+9}}{2}\le y\le\dfrac{-m-5+\sqrt{9m^2+14m+9}}{2}\)

\(\Rightarrow y_{min}=\dfrac{-m-5-\sqrt{9m^2+14m+9}}{2}\le3\)

\(\Rightarrow\sqrt{9m^2+14m+9}\ge-m-11\)

BPT này đúng với mọi m. Vậy bài toán thỏa mãn với mọi m

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!