Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
2b)\,\,Xet\,\,pthdgd:\,\,kx + 1 - k = - {x^2} + 3x - 2 \Leftrightarrow {x^2} + \left( {k - 3} \right)x + 3 - k = 0\,\,\left( {**} \right)\\
y = kx + 1 - k\,\,\,cat\,\,\left( P \right)\,\,tai\,\,2\,\,diem\,\,pb \Rightarrow \left( {**} \right)\,\,co\,\,2\,\,nghiem\,\,pb\\
\Leftrightarrow \Delta = {\left( {k - 3} \right)^2} - 4.\left( {3 - k} \right) > 0 \Leftrightarrow {k^2} - 2k - 3 > 0 \Leftrightarrow \left[ \begin{array}{l}
k > 3\\
k < - 1
\end{array} \right.\\
c)\,\,Voi\,\,\left[ \begin{array}{l}
k > 3\\
k < - 1
\end{array} \right.,\,\,gia\,\,su\,\,\left( {**} \right)\,\,co\,\,2\,\,\,\,nghiem\,\,pb\,\,{x_1},\,\,{x_2} \Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = 3 - k\\
{x_1}{x_2} = 3 - k
\end{array} \right.\,\,\left( {DL\,\,Vi - et} \right)\\
\Rightarrow A\left( {{x_1};k{x_1} + 1 - k} \right);\,\,B\left( {{x_2};k{x_2} + 1 - k} \right)\\
AB = 2 \Leftrightarrow A{B^2} = 4\\
\Rightarrow {\left( {{x_2} - {x_1}} \right)^2} + {k^2}{\left( {{x_2} - {x_1}} \right)^2} = 4\\
\Leftrightarrow {\left( {{x_2} - {x_1}} \right)^2}\left( {1 + {k^2}} \right) = 4\\
\Leftrightarrow \left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \right]\left( {1 + {k^2}} \right) = 4\\
\Leftrightarrow \left[ {{{\left( {3 - k} \right)}^2} - 4\left( {3 - k} \right)} \right]\left( {1 + {k^2}} \right) = 4\\
\Leftrightarrow \left( {{k^2} - 2k - 3} \right)\left( {1 + {k^2}} \right) = 4\\
\Leftrightarrow {k^2} - 2k - 3 + {k^4} - 2{k^3} - 3{k^2} - 4 = 0\\
\Leftrightarrow {k^4} - 2{k^3} - 2{k^2} - 2k - 7 = 0\\
\Leftrightarrow ...
\end{array}\)
