\(A^2=\frac{4\left(m+1\right)^2}{m^2+3}\)
\(\Rightarrow A^2\left(m^2+3\right)-4\left(m+1\right)^2=0\)
\(\Leftrightarrow A^2m^2+3A^2-4m^2-8m-4=0\)
\(\Leftrightarrow\left(A^2-4\right)m^2-8m+3A^2-4=0\)
*Với \(A=\pm2\):
\(\Rightarrow8m+12-4=0\)
\(\Leftrightarrow m=-1\)
*Với \(A\ne\pm2:\)
Đk để pt có ng0 thì \(\Delta\ge0\)
\(\Rightarrow64-4\left(A^2-4\right)\left(3A^2-4\right)\ge0\)
\(\Leftrightarrow16-3A^4+16A^2-16\ge0\)
\(\Leftrightarrow0\le A^2\le\frac{16}{3}\)
\(\Rightarrow A^2_{max}=\frac{16}{3}\Leftrightarrow A=\pm\sqrt{\frac{16}{3}}=\pm\frac{4\sqrt{3}}{3}\)
Vậy Amax\(=\frac{4\sqrt{3}}{3}\Leftrightarrow\frac{2\left|m+1\right|}{\sqrt{m^2+3}}=\frac{4\sqrt{3}}{3}\)
*Với \(m\ge-1\)
\(\Rightarrow\frac{m+1}{\sqrt{m^2+3}}=\frac{2\sqrt{3}}{3}\)
\(\Rightarrow3m+3=\sqrt{12\left(m^2+3\right)}\)
\(\Leftrightarrow9m^2+9+18m=12m^2+36\)
\(\Leftrightarrow3m^2-18m+27=0\)
\(\Leftrightarrow m=3\left(TM\right)\)
*Với m<1:
\(\Rightarrow-3m-3=\sqrt{12\left(m^2+3\right)}\)
\(\Leftrightarrow3m^2-18m+27=0\)
\(\Leftrightarrow m=3\left(KTM\right)\)
Vậy Amax\(=\frac{4\sqrt{3}}{3}\Leftrightarrow m=3\)
\(A=2\sqrt{\frac{\left(m+1\right)^2}{m^2+3}}\)
Quy về bài toán tìm max của \(P=\frac{m^2+2m+1}{m^2+3}\)
\(P=1+\frac{2m-2}{m^2+3}\)
Đặt \(B=\frac{2m-2}{m^2+3}\)
\(\Rightarrow Bm^2-2m+3B+2=0\)(*)
Xét B=0=>m=1
Xét \(B\ne0\)\(\Rightarrow\)Để (*) có nghiệm thì \(\Delta\)'=\(1-3B^2-2B\ge0\)
\(\Leftrightarrow\left(1-3B\right)\left(1+B\right)\ge0\)
\(\Leftrightarrow-1\le B\le\frac{1}{3}\)
\(\Rightarrow A\le\)\(\frac{4}{\sqrt{3}}\)
"="<=>m=3
