Pt:(m+ 1)x^2-2(m+1)x-m+2=0 e) có no âm pb f) có no x1,x2 thỏa |x1- x2| bé hơn bằng 3 29/11/2021 Bởi Ximena Pt:(m+ 1)x^2-2(m+1)x-m+2=0 e) có no âm pb f) có no x1,x2 thỏa |x1- x2| bé hơn bằng 3
Đáp án: f. \(m \in \left( { – \infty ; – 13} \right] \cup \left( {1; + \infty } \right)\) Giải thích các bước giải: \(e.\left\{ \begin{array}{l}{m^2} + 2m + 1 – \left( {m + 1} \right)\left( {2 – m} \right) > 0\\\frac{{2m + 2}}{{m + 1}} = 2 < 0\left( {voly} \right)\\\frac{{2 – m}}{{m + 1}} > 0\end{array} \right.\) ⇒ Không tồn tại m để pt có 2 nghiệm âm phân biệt \(\begin{array}{l}f.\left\{ \begin{array}{l}{m^2} + 2m + 1 – \left( {m + 1} \right)\left( {2 – m} \right) > 0\\0 \le {x_1}^2 – 2{x_1}{x_2} + {x_2}^2 \le 9\end{array} \right.\\ \to \left\{ \begin{array}{l}2{m^2} – m – 1 > 0\\0 \le \left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2} \right) – 4{x_1}{x_2} \le 9\end{array} \right.\\ \to \left\{ \begin{array}{l}m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\0 \le {\left( {{x_1} + {x_2}} \right)^2} – 4{x_1}{x_2} \le 9\end{array} \right.\\ \to \left\{ \begin{array}{l}m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\{2^2} – 4.\frac{{2 – m}}{{m + 1}} \ge 0\\{2^2} – 4.\frac{{2 – m}}{{m + 1}} \le 9\end{array} \right.\\ \to \left\{ \begin{array}{l}m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\\frac{{m + 1 – 2 + m}}{{m + 1}} \ge 0\\\frac{{4m + 4 – 8 + 4m – 9m – 9}}{{m + 1}} \le 0\end{array} \right.\\ \to \left\{ \begin{array}{l}m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\\frac{{2m – 1}}{{m + 1}} \ge 0\\\frac{{ – m – 13}}{{m + 1}} \le 0\end{array} \right.\\ \to \left\{ \begin{array}{l}m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\m \in \left( { – \infty ; – 1} \right] \cup \left[ {\frac{1}{2}; + \infty } \right)\\m \in \left( { – \infty ; – 13} \right] \cup \left[ { – 1; + \infty } \right)\end{array} \right.\\KL:m \in \left( { – \infty ; – 13} \right] \cup \left( {1; + \infty } \right)\end{array}\) Bình luận
Đáp án:
f. \(m \in \left( { – \infty ; – 13} \right] \cup \left( {1; + \infty } \right)\)
Giải thích các bước giải:
\(e.\left\{ \begin{array}{l}
{m^2} + 2m + 1 – \left( {m + 1} \right)\left( {2 – m} \right) > 0\\
\frac{{2m + 2}}{{m + 1}} = 2 < 0\left( {voly} \right)\\
\frac{{2 – m}}{{m + 1}} > 0
\end{array} \right.\)
⇒ Không tồn tại m để pt có 2 nghiệm âm phân biệt
\(\begin{array}{l}
f.\left\{ \begin{array}{l}
{m^2} + 2m + 1 – \left( {m + 1} \right)\left( {2 – m} \right) > 0\\
0 \le {x_1}^2 – 2{x_1}{x_2} + {x_2}^2 \le 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2{m^2} – m – 1 > 0\\
0 \le \left( {{x_1}^2 + 2{x_1}{x_2} + {x_2}^2} \right) – 4{x_1}{x_2} \le 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\
0 \le {\left( {{x_1} + {x_2}} \right)^2} – 4{x_1}{x_2} \le 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\
{2^2} – 4.\frac{{2 – m}}{{m + 1}} \ge 0\\
{2^2} – 4.\frac{{2 – m}}{{m + 1}} \le 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\
\frac{{m + 1 – 2 + m}}{{m + 1}} \ge 0\\
\frac{{4m + 4 – 8 + 4m – 9m – 9}}{{m + 1}} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\
\frac{{2m – 1}}{{m + 1}} \ge 0\\
\frac{{ – m – 13}}{{m + 1}} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { – \infty ; – \frac{1}{2}} \right) \cup \left( {1; + \infty } \right)\\
m \in \left( { – \infty ; – 1} \right] \cup \left[ {\frac{1}{2}; + \infty } \right)\\
m \in \left( { – \infty ; – 13} \right] \cup \left[ { – 1; + \infty } \right)
\end{array} \right.\\
KL:m \in \left( { – \infty ; – 13} \right] \cup \left( {1; + \infty } \right)
\end{array}\)