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a) \(\left(x^2-2\right)\left(k-1\right)x+2k-5=0\)
\(\Delta=\left(k-1\right)^2-2k+5\)
\(=k^2-4x+6=\left(k-2\right)^2+2>0\)
=> PT luôn có nghiệm với mọi k
a)x2+5x+3m-1
- Pt có 2 nghiệm trái dấu khi
\(\Delta>0\Leftrightarrow m< \frac{29}{12}\).pt có 2 nghiệm phân biệt
\(x_{1,2}=\frac{5\pm\sqrt{29-12m}}{2}\)
- Pt có 2 nghiệm âm phân biệt khi
\(\begin{cases}\Delta\ge0\\p=1\end{cases}\)\(\Leftrightarrow\begin{cases}29-12m\ge0\\3m-1=1\end{cases}\)\(\Leftrightarrow m=\frac{2}{3}\left(tm\right)\)
- Pt có 2 nghiệm dương phân biệt khi
\(\begin{cases}\Delta>0\\p=\frac{c}{a}>0\\S=\frac{b}{a}>0\end{cases}\)\(\Leftrightarrow\begin{cases}29-12m>0\\3m-1>0\\5>0\left(\text{đúng}\right)\end{cases}\)\(\Leftrightarrow\frac{1}{3}< m< \frac{29}{12}\)
Điều kiện để có pt bậc hai có 2 nghiệm phân biệt cùng dấu là:
\(\hept{\begin{cases}\Delta'>0\\x_1.x_2=\frac{c}{a}>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}k^2-4k+5>0\\4k-5>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(k-2\right)^2+1>0\\k>\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow k>\frac{5}{4}\)
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