Question

bài 10 tìm tất các giá trị của tham số m để phương trình sau có hai nghiệm cùng dấu

a) x²+(2m-2)x+m+1=0

b)-x²+(m-2)x+2m-1=0

c) x²+mx+m-3/4=0

d)4x²+4(2m-1)x+m=0

e)x²-(m+1)x+m-1=0

f)4x²+2(m-1)x+m-1=0

g)(m-2)x²-2(m-2)x+1=0

h)(m-2)x²+2(2m-3)x+5m-6=0

Answer 1

Để pt có 2 nghiệm cùng dấu: \(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\\Delta\ge0\\x_1x_2=\frac{c}{a}>0\end{matrix}\right.\)

a/ \(\left\{{}\begin{matrix}\Delta'=\left(m-1\right)^2-m-1\ge0\\m+1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-3m\ge0\\m>-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m\ge3\\-1< m\le0\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}\Delta=\left(m-2\right)^2-4\left(2m-1\right)\ge0\\\frac{2m-1}{-1}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-12m+8\ge0\\m< \frac{1}{2}\end{matrix}\right.\) \(\Rightarrow m< \frac{1}{2}\)

c/ \(\left\{{}\begin{matrix}\Delta=m^2-4\left(m-\frac{3}{4}\right)\ge0\\m-\frac{3}{4}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-4m+3\ge0\\m>\frac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< m\le1\\m\ge3\end{matrix}\right.\)

Answer 2

d/

\(\left\{{}\begin{matrix}\Delta'=4\left(2m-1\right)^2-4m\ge0\\\frac{m}{4}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-5m+1\ge0\\m>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}0< m< \frac{1}{4}\\m>1\end{matrix}\right.\)

e/

\(\left\{{}\begin{matrix}\Delta=\left(m+1\right)^2-4\left(m-1\right)\ge0\\m-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-2m+5\ge0\\m>1\end{matrix}\right.\) \(\Rightarrow m>1\)

f/

\(\left\{{}\begin{matrix}\Delta'=\left(m-1\right)^2-4\left(m-1\right)\ge0\\\frac{m-1}{4}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-6m+5\ge0\\m>1\end{matrix}\right.\) \(\Rightarrow m\ge5\)

Answer 3

g/

\(\left\{{}\begin{matrix}m-2\ne0\\\Delta'=\left(m-2\right)^2-\left(m-2\right)\ge0\\\frac{1}{m-2}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\\left(m-2\right)\left(m-3\right)\ge0\\m>2\end{matrix}\right.\)

\(\Rightarrow m\ge3\)

h/

\(\left\{{}\begin{matrix}m-2\ne0\\\Delta'=\left(2m-3\right)^2-\left(m-2\right)\left(5m-6\right)\ge0\\\frac{5m-6}{m-2}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\-m^2+4m-3\ge0\\\left[{}\begin{matrix}m>2\\m< \frac{6}{5}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}1\le m< \frac{6}{5}\\2< m\le3\end{matrix}\right.\)