a: Khi m=2 thì phương trình sẽ trở thành:
\(x^2-2\left(2-1\right)x-2\cdot2-1=0\)
=>\(x^2-2x-5=0\)
=>\(x^2-2x+1=6\)
=>\(\left(x-1\right)^2=6\)
=>\(x-1=\pm\sqrt{6}\)
=>\(x=1\pm\sqrt{6}\)
b: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(-2m-1\right)\)
\(=\left(2m-2\right)^2+4\left(2m+1\right)\)
\(=4m^2-8m+4+8m+4=4m^2+8>0\) với mọi m
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)=2m-2\\x_1x_2=\dfrac{c}{a}=-2m-1\end{matrix}\right.\)
\(2x_1+3x_2+3x_1x_2=-11\)
=>\(2x_1+3x_2+3\left(-2m-1\right)=-11\)
=>\(2x_1+3x_2=-11-3\cdot\left(-2m-1\right)=-11+6m+3=6m-8\)
Do đó, ta có hệ:
\(\left\{{}\begin{matrix}x_1+x_2=2m-2\\2x_1+3x_2=6m-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x_1+2x_2=4m-4\\2x_1+3x_2=6m-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x_2=4m-4-6m+8\\x_1+x_2=2m-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x_2=2m-4\\x_1=2m-2-\left(2m-4\right)=2\end{matrix}\right.\)
\(x_1\cdot x_2=-2m-1\)
=>-2m-1=2(2m-4)
=>4m-8=-2m-1
=>6m=7
=>\(m=\dfrac{7}{6}\)
a)
Với m = 2 ta có PT: \(x^2-2x-5=0\)
\(\Delta=\left(-2\right)^2-4\cdot1\cdot\left(-5\right)=4+20=24>0\)
\(\Rightarrow\) PT có 2 nghiệm phân biệt
\(x_1=\dfrac{2+\sqrt{24}}{2}=\dfrac{2+2\sqrt{6}}{2}=1+\sqrt{6}\)
\(x_2=\dfrac{2-\sqrt{24}}{2}=\dfrac{2-2\sqrt{6}}{2}=1-\sqrt{6}\)
b)
Theo Vi-ét ta có:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\left(1\right)\\x_1\cdot x_2=-2m-1\left(2\right)\end{matrix}\right.\)
Theo đề bài ta có: \(2x_1+3x_2+3x_1x_2=-11\Leftrightarrow2x_1+3x_2=-3\left(-2m-1\right)-11\left(3\right)\)
Từ (1)(3) ta có:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\2x_1+3x_2=6m+3-11=6m-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_1+3x_2=6m-6\\2x_1+3x_2=6m-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=6m-6-6m+8=2\\x_1+x_2=2m-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=2\\2+x_2=2m-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=2\\x_2=2m-4\end{matrix}\right.\)
Thay x1,x2 vào (2) ta có:
\(2\cdot\left(2m-4\right)=-2m-1\)
\(\Leftrightarrow4m-8=-2m-1\)
\(\Leftrightarrow6m=7\)
\(\Leftrightarrow m=\dfrac{7}{6}\)
a. Em tự giải
b.
\(\Delta'=\left(m-1\right)^2+2m+1=m^2+2>0;\forall m\)
Pt luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-2m-1\end{matrix}\right.\)
\(2x_1+3x_2+3x_1x_2=-11\)
\(\Leftrightarrow2\left(x_1+x_2\right)+3x_1x_2+x_2=-11\)
\(\Leftrightarrow4\left(m-1\right)+3\left(-2m-1\right)+x_2=-11\)
\(\Leftrightarrow x_2=2m-4\)
\(\Rightarrow x_1=2\left(m-1\right)-x_2=2\)
Thế vào \(x_1x_2=-2m-1\)
\(\Rightarrow2\left(2m-4\right)=-2m-1\)
\(\Rightarrow m=\dfrac{7}{6}\)
