Nếu \(m=-1\)
\(\left(1\right)\Leftrightarrow-2x+2=0\Leftrightarrow x=1\)
Nếu \(m\ne-1\)
a, Phương trình có hai nghiệm phân biệt khi \(\Delta'=\left(m+2\right)^2-\left(m+1\right)\left(m+3\right)=7>0\)
\(\Rightarrow\) Phương trình luôn có hai nghiệm phân biệt \(x_1;x_2\)
Yêu cầu bài toán thỏa mãn khi phương trình có hai nghiệm thỏa mãn \(x_1>x_2>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+1\right).f\left(0\right)>0\\\frac{m+2}{m+1}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m+1\right)\left(m+3\right)>0\\\frac{m+2}{m+1}>0\end{matrix}\right.\)
\(\Leftrightarrow m< -3;m>-1\)
b, Theo định lí Vi-ét \(\left\{{}\begin{matrix}x_1x_2=\frac{m+3}{m+1}\\x_1+x_2=\frac{2\left(m+2\right)}{m+1}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2_2=\frac{m+3}{m+1}\left(1\right)\\3x_2=\frac{2\left(m+2\right)}{m+1}\left(2\right)\end{matrix}\right.\)
Trừ vế \(\left(2\right)\) cho \(\left(1\right)\) ta được \(3x_2-2x^2_2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x_2=1\\x_2=\frac{1}{2}\end{matrix}\right.\)
Nếu \(x_2=1\Rightarrow\left\{{}\begin{matrix}m+3=2m+2\\2m+4=3m+3\end{matrix}\right.\Leftrightarrow m=1\)
Nếu \(x_2=\frac{1}{2}\Rightarrow\left\{{}\begin{matrix}2m+6=m+1\\4m+8=3m+3\end{matrix}\right.\Leftrightarrow m=-5\)
c, Phương trình có hai nghiệm trái dấu khi \(\left(m+1\right)f\left(0\right)< 0\)
\(\Leftrightarrow\left(m+1\right)\left(m+3\right)< 0\)
\(\Leftrightarrow-3< m< -1\)
d, \(x_1^2+x_2^2=11-3x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=11-3x_1x_2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=11\)
\(\Leftrightarrow\left(\frac{2m+4}{m+1}\right)^2+\frac{m+3}{m+1}=11\)
\(\Leftrightarrow\frac{4m^2+16m+16}{\left(m+1\right)^2}+\frac{\left(m+3\right)\left(m+1\right)}{\left(m+1\right)^2}=11\)
\(\Leftrightarrow4m^2+16m+16+\left(m+3\right)\left(m+1\right)=11\left(m+1\right)^2\)
\(\Leftrightarrow5m^2+20m+19=11m^2+22m+11\)
\(\Leftrightarrow6m^2+2m-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-\frac{4}{3}\end{matrix}\right.\)
