xét pt \(x^2+5x+3m-1=0\)
có \(\Delta=5^2-4\left(3m-1\right)=25-12m+4=29-12m\)
để pt đã cho có 2 nghiệm \(x_1,x_2\) thì \(\Delta\ge0\)
\(\Leftrightarrow29-12m\ge0\)
\(\Leftrightarrow m\le\dfrac{29}{12}\)
ta có vi - ét \(\left\{{}\begin{matrix}x_1+x_2=-5\left(1\right)\\x_1.x_2=3m-1\left(2\right)\end{matrix}\right.\)
theo bài ra ta có \(x_1^3-x_2^3+3x_1x_2=75\)
\(\Leftrightarrow x_1^3+x_2^3-2x_2^3+3x_1x_2-75=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1.x_2\left(x_1+x_2\right)+3x_1x_2-2x_3^3-75=0\)
\(\Leftrightarrow\left(-5\right)^3-3\left(3m-1+1\right)-2x_3^3-75=0\)
\(\Leftrightarrow125-9m-2x_3^3-75=0\)
\(\Leftrightarrow-2x_3^3-9m+50=0\)
đến chỗ này mình bí rồi 
Theo hệ thức Vi-et
\(x_1+x_2=-5\)
\(x_1x_2=3m-1\)
Ta có:
\(x_1^3+x_2^3+3x_1x_2=75\)
\(\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+3x_1x_2=75\)
\(\Leftrightarrow\left(-5\right)^3-3\left(3m-1\right)\left(-5\right)+3\left(3m+1\right)=75\)
\(\Leftrightarrow-125+15\left(3m-1\right)+9m+3-75=0\)
\(\Rightarrow-197+45m-15+9m=0\)
\(\Leftrightarrow54m=212\)
\(\Leftrightarrow m=\dfrac{106}{27}\)
\(\)
a) Với m=2
\(\left\{{}\begin{matrix}x-2y=5\\2x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x-2y=5\\mx-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5+2y\\m\left(5+2y\right)-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5+2y\\5m+2ym-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5+2y\\2ym-m=4-5m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5+2y\\m\left(2y-1\right)=4-5m\end{matrix}\right.\)
Để hpt có no duy nhất thì m\(\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}x=5+2y\\y=\frac{4+4m}{2m}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{4+9m}{m}\\y=\frac{2+2m}{m}\end{matrix}\right.\)
c) \(\frac{4+9m}{m}=\left|\frac{2+2m}{m}\right|\)
TH1
\(\frac{4+9m}{m}=\frac{2+2m}{m}\)
--> \(\frac{2+7m}{m}=0\) .Do m #0
-->2+7m=0-->7m=-1-->m=-1/7
TH2
\(\frac{4+9m}{m}=\frac{-2-2m}{m}\)
-->\(\frac{6+11m}{m}=0\).Do m #0
--> 6+11m=0--->11m=-6--->-6/11
Gửi nhờ cho bn mk nhé
