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a/ Bạn tự giải
b/ \(\Delta=9-4\left(m-2\right)=17-4m>0\Rightarrow m< \frac{17}{4}\)
Khi đó theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=m-2\end{matrix}\right.\)
\(x^3_1-x_2^3+9x_1x_2=81\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(\left(x_1+x_2\right)^2-x_1x_2\right)+9x_1x_2=81\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(11-m\right)+9m-18=81\)
\(\Rightarrow x_1-x_2=\frac{99-9m}{11-m}=9\)
Ta có hệ: \(\left\{{}\begin{matrix}x_1-x_2=9\\x_1+x_2=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=6\\x_2=-3\end{matrix}\right.\)
\(\Rightarrow m=x_1x_2+2=-16\)
\(x^2m-2\left(m-1\right)x+m+1=0\)
\(\Delta=b^2-4ac\)
\(\Rightarrow\Delta=4m+4\)
Để phương trình có 2 nghiệm phân biệt
\(\Rightarrow\Delta>0\Leftrightarrow m>-1\)
Theo định lý Viet
\(\Rightarrow\hept{\begin{cases}x_1+x_2=\frac{-b}{a}\\x_1x_2=\frac{c}{a}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1+x_2=\frac{2\left(m-1\right)}{m}\\x_1.x_2=\frac{m+1}{m}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(x_1+x_2\right)^2=\left[\frac{2\left(m-1\right)}{m}\right]^2\\2x_1x_2=\frac{2\left(m+1\right)}{m}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x_1^2+x_2^2+2x_1x_2=\frac{4\left(m-1\right)^2}{m^2}\left(1\right)\\2x_1x_2=\frac{2\left(m+1\right)}{m}\end{cases}}\)
Xét phương trình ( 1 )
\(pt\left(1\right)\Leftrightarrow16+\frac{2\left(m+1\right)}{m}=\frac{4\left(m-1\right)^2}{m^2}\)
\(\Leftrightarrow\frac{16m+2\left(m+1\right)}{m}=\frac{4\left(m-1\right)^2}{m^2}\)
\(\Leftrightarrow\frac{18m+2}{m}=\frac{4\left(m^2-2m+1\right)}{m^2}\)
\(\Leftrightarrow m^2\left(18m+2\right)=4m\left(m^2-2m+1\right)\)với m khác 0
\(\Leftrightarrow m\left(18m+2\right)=4\left(m^2-2m+1\right)\)
\(\Leftrightarrow18m^2+2m=4m^2-8m+4\)
\(\Leftrightarrow14m^2+10m-4=0\)
\(\Delta=b^2-4ac\)
\(\Rightarrow\Delta=324\)
\(\Rightarrow\hept{\begin{cases}m_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-10+\sqrt{324}}{28}\\m_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-10-\sqrt{324}}{28}\end{cases}}\)
Do \(m>-1\)
\(\Rightarrow m=\frac{-10+\sqrt{324}}{28}\)
a, \(m=-8=>x^2-3x-10=0\)
\(\Delta=\left(-3\right)^2-4\left(-10\right)=49>0\)
=>pt có 2 nghiệm phân biệt \(=>\left[{}\begin{matrix}x1=\dfrac{3+\sqrt{49}}{2}=5\\x2=\dfrac{3-\sqrt{49}}{2}=-2\end{matrix}\right.\)
b, pt(1) \(=>\Delta=\left(-3\right)^2-4\left(m-2\right)=9-4m+8=17-4m\)
pt (1) có 2 nghiệm phân biệt x1,x2 khi \(17-4m>0< =>m< \dfrac{17}{4}\)
theo vi ét \(=>\left\{{}\begin{matrix}x1+x2=3\left(1\right)\\x1x2=m-2\end{matrix}\right.\)
\(x1^3-x2^3+9x1x2=81\)
\(=>\left(x1-x2\right)\left(x1^2+x1x2+x2^2\right)+9\left(m-2\right)=81\)
\(=>x1-x2=\dfrac{81-9\left(m-2\right)}{\left[\left(x1+x2\right)^2-x1x2\right]}\)
\(=>x1-x2=\dfrac{99-9m}{\left[3^2-m+2\right]}=\dfrac{99-9m}{11-m}=9\left(2\right)\)
từ (1)(2)=> hệ pt: \(\left\{{}\begin{matrix}x1+x2=3\\x1-x2=9\end{matrix}\right.=>\left\{{}\begin{matrix}x1=6\\x2=-3\end{matrix}\right.\)
\(=>x1x2=6.\left(-3\right)=m-2=>m=-16\left(tm\right)\)