Câu 2:
\(a,\Leftrightarrow\Delta'=\left(1-m\right)^2-\left(m^2-m\right)>0\\ \Leftrightarrow m^2-2m+1-m^2+m>0\\ \Leftrightarrow1-m>0\Leftrightarrow m< 1\\ b,\text{Áp dụng Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(1-m\right)\\x_1x_2=m^2-m\end{matrix}\right.\\ \left(2x_1-1\right)\left(2x_2-1\right)-x_1x_2=1\\ \Leftrightarrow2x_1x_2-2\left(x_1+x_2\right)+1-x_1x_2=1\\ \Leftrightarrow x_1x_2-2\left(x_1+x_2\right)=0\\ \Leftrightarrow m^2-m-4\left(1-m\right)=0\\ \Leftrightarrow m^2+3m-4=0\\ \Leftrightarrow\left(m-1\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-4\left(tm\right)\end{matrix}\right.\)
Vậy m=-4
Câu 1:
\(1,\Leftrightarrow2x-2=3\Leftrightarrow x=\dfrac{5}{2}\\ 2,ĐK:x\ne\pm1\\ PT\Leftrightarrow\dfrac{2x^2+2x-1}{x^2-1}=2\\ \Leftrightarrow2x^2+2x-1=2x^2-2\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\\ 3,\Leftrightarrow\left[{}\begin{matrix}3x-2=2x-1\\3x-2=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\)
\(4,\Leftrightarrow\left[{}\begin{matrix}3x-1=2-x\left(x\ge\dfrac{1}{3}\right)\\3x-1=x-2\left(x< \dfrac{1}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(tm\right)\\x=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\\ 5,\Leftrightarrow4x^2-2x+10=9x^2-6x+1\left(x\le\dfrac{1}{3}\right)\\ \Leftrightarrow5x^2-4x-9=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(6,\Leftrightarrow3x^2-9x+1=x^2-4x+4\left(x\ge2\right)\\ \Leftrightarrow2x^2-5x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ 7,\Leftrightarrow2x^2+3x-4=7x+2\left(x\ge-\dfrac{2}{7}\right)\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
