ĐK: m\(\ne0\)
\(\Delta=4\left(m-4\right)^2-4\cdot m\cdot\left(m+7\right)=4m^2-32m+64-4m^2-28m=-60m+64\)
PT có nghiệm\(\Leftrightarrow-60m+64\ge0\Leftrightarrow m\le\dfrac{16}{15}\)
Theo Vi-ét và đề bài ta có:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2\left(m-4\right)}{m}\left(1\right)\\x_1\cdot x_2=\dfrac{m+7}{m}\left(2\right)\\x_1-2x_2=0\left(3\right)\end{matrix}\right.\)
Từ (1)(3) ta có:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2\left(m-4\right)}{m}\\x_1-2x_2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_2=\dfrac{-2\left(m-4\right)}{m}\\x_1-2_2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-2\left(m-4\right)}{3m}\\x_1-2\left(\dfrac{-2\left(m-4\right)}{3m}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=\dfrac{-2\left(m-4\right)}{3m}\\x_1=\dfrac{-4\left(m-4\right)}{3m}\end{matrix}\right.\)
Thay x1, x2 vào (2) ta có:
\(\dfrac{-2\left(m-4\right)}{3m}\cdot\dfrac{-4\left(m-4\right)}{3m}=\dfrac{m+7}{m}\)
\(\Leftrightarrow\dfrac{8\left(m-4\right)^2}{9m^2}=\dfrac{m+7}{m}\)
\(\Leftrightarrow8m\left(m-4\right)^2=9m^2\left(m+7\right)\)
\(\Leftrightarrow8m^3-64m^2+128m=9m^3+7m^2\)
\(\Leftrightarrow8m^3-64m^2+128m-9m^3-7m^2=0\)
\(\Leftrightarrow-m^3-71m^2+128m=0\)
\(\Leftrightarrow-m\left(m^2+71m-128\right)=0\)
Vì \(m^2+71m-128=m^2+2\cdot\dfrac{71}{2}m+\dfrac{5041}{4}-\dfrac{5553}{4}=\left(m+\dfrac{71}{2}\right)^2-\dfrac{5553}{4}\ne0\)
\(\Rightarrow-m=0\Rightarrow m=0\) (Loại)
\(\Rightarrow\) PT Vô nghiệm
\(\Rightarrow\) Ko có GT nào của m thỏa mãn
TH1: m=0
Phương trình sẽ trở thành:
\(0x^2+2\left(0-4\right)x+0+7=0\)
=>-8x+7=0
=>-8x=-7
=>\(x=\dfrac{7}{8}\)
=>Phương trình chỉ có 1 nghiệm duy nhất
=>Loại
TH2: \(m\ne0\)
\(\text{Δ}=\left[2\left(m-4\right)\right]^2-4m\left(m+7\right)\)
\(=\left(2m-8\right)^2-4m\left(m+7\right)\)
\(=4m^2-32m+64-4m^2-28m=-60m+64\)
Để phương trình có hai nghiệm thì Δ>=0
=>-60m+64>=0
=>-60m>=-64
=>\(m< =\dfrac{16}{15}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-2\left(m-4\right)}{m}=\dfrac{-2m+8}{m}\\x_1x_2=\dfrac{c}{a}=\dfrac{m+7}{m}\end{matrix}\right.\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x_1-2x_2=0\\x_1+x_2=\dfrac{-2m+8}{m}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=2x_2\\3x_2=\dfrac{-2m+8}{m}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_2=\dfrac{-2m+8}{3m}\\x_1=\dfrac{2\cdot\left(-2m+8\right)}{3m}=\dfrac{-4m+16}{3m}\end{matrix}\right.\)
\(x_1\cdot x_2=\dfrac{m+7}{m}\)
=>\(\dfrac{\left(-2m+8\right)\left(-4m+16\right)}{9m^2}=\dfrac{m+7}{m}\)
=>\(\dfrac{\left(2m-8\right)\left(4m-16\right)}{9m^2}=\dfrac{9m\left(m+7\right)}{9m^2}\)
=>\(9m\left(m+7\right)=\left(2m-8\right)\left(4m-16\right)\)
=>\(9m^2+63m=8m^2-64m+128\)
=>\(m^2+127m-128=0\)
=>(m+128)(m-1)=0
=>\(\left[{}\begin{matrix}m=-128\left(nhận\right)\\m=1\left(nhận\right)\end{matrix}\right.\)
