Phương trình có hai nghiệm âm phân biệt hay dương phân biệt bạn?

Hay hai nghiệm trái dấu?

Để pt có 2 nghiệm dương (ko yêu cầu pb?) \(\left\{{}\begin{matrix}a\ne0\\\Delta\ge0\\x_1+x_2=-\frac{b}{a}>0\\x_1x_2=\frac{c}{a}>0\end{matrix}\right.\)

a/ \(\left\{{}\begin{matrix}\Delta=\left(2m-1\right)^2+4m-4\ge0\\x_1+x_2=2m+1>0\\x_1x_2=-m+1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-3\ge0\\m>-\frac{1}{2}\\m< 1\end{matrix}\right.\) \(\Rightarrow\frac{\sqrt{3}}{2}\le m< 1\)

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

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

c/

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

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-2m\ge0\\m< -1\\m>-\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\) không tồn tại m thỏa mãn

d/

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

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-8m+1\ge0\\m>\frac{1}{2}\\m>0\end{matrix}\right.\) \(\Rightarrow m\ge\frac{2+\sqrt{3}}{2}\)

e/

\(\left\{{}\begin{matrix}\Delta=\left(m+1\right)^2-4m\ge0\\x_1+x_2=m+1>0\\x_1x_2=m>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)^2\ge0\\m>-1\\m>0\end{matrix}\right.\) \(\Rightarrow m>0\)

f/

\(\left\{{}\begin{matrix}m-2\ne0\\\Delta'=\left(2m-3\right)^2-\left(m-2\right)\left(5m-6\right)\ge0\\x_1+x_2=\frac{2\left(3-2m\right)}{m-2}>0\\x_1x_2=\frac{5m-6}{m-2}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\-m^2+4m-3\ge0\\\frac{3-2m}{m-2}>0\\\frac{5m-6}{m-2}>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\ne2\\1\le m\le3\\\frac{3}{2}< m< 2\\\left[{}\begin{matrix}m< \frac{6}{5}\\m>2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại m thỏa mãn

Để pt có 2 nghiệm âm (không cần phân biệt) \(\left\{{}\begin{matrix}a\ne0\\\Delta\ge0\\x_1+x_2=-\frac{b}{a}< 0\\x_1x_2=\frac{c}{a}>0\end{matrix}\right.\)

a/

\(\left\{{}\begin{matrix}\Delta=\left(2m-1\right)^2-4\left(m+1\right)\ge0\\x_1+x_2=-2m+1< 0\\x_1x_2=m+1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-8m-3\ge0\\m>\frac{1}{2}\\m>-1\end{matrix}\right.\) \(\Rightarrow m\ge\frac{2+\sqrt{7}}{2}\)

b/

\(\left\{{}\begin{matrix}\Delta=\left(m-2\right)^2+4\left(2m-1\right)\ge0\\x_1+x_2=m-2< 0\\x_1x_2=1-2m>0\end{matrix}\right.\)

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

c/

\(\left\{{}\begin{matrix}\Delta=m^2-4\left(m-\frac{3}{4}\right)\ge0\\x_1+x_2=-m< 0\\x_1x_2=m-\frac{3}{4}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-4m+3\ge0\\m>0\\m>\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m\ge3\\\frac{3}{4}< m\le1\end{matrix}\right.\)

d/

\(\left\{{}\begin{matrix}\Delta'=4\left(2m-1\right)^2-4m\ge0\\x_1+x_2=1-2m< 0\\x_1x_2=\frac{m}{4}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-5m+1\ge0\\m>\frac{1}{2}\\m>0\end{matrix}\right.\) \(\Rightarrow m\ge1\)

e/

\(\left\{{}\begin{matrix}\Delta=\left(m+1\right)^2-4\left(m-1\right)\ge0\\x_1+x_2=m+1< 0\\x_1x_2=m-1>0\end{matrix}\right.\)

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

\(\Rightarrow\) Không tồn tại m thỏa mãn

f/

\(\left\{{}\begin{matrix}m-2\ne0\\\Delta'=\left(m-2\right)^2-\left(m-2\right)\ge0\\x_1+x_2=2< 0\left(vô-lý\right)\\x_1x_2=\frac{1}{m-2}>0\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại m thỏa mãn

a: Khi m=1 thì (1): x^2-2(1-2)x+1^2-5-4=0

=>x^2+2x-8=0

=>(x+4)(x-2)=0

=>x=2 hoặc x=-4

b: Δ=(2m-4)^2-4(m^2-5m-4)

=4m^2-16m+16-4m^2+20m+16

=4m+32

Để pt có hai nghiệm phân biệt thì 4m+32>0

=>m>-8

x1^2+x2^2=-3x1x2-4

=>(x1+x2)^2+x1x2+4=0

=>(2m-4)^2+m^2-5m-4+4=0

=>4m^2-16m+16+m^2-5m=0

=>5m^2-21m+16=0

=>(m-1)(5m-16)=0

=>m=16/5 hoặc m=1

a: Khi m=-8 thì (1) sẽ là x^2+6x=0

=>x=0; x=-6

\(\Delta'=\left(m+1\right)^2-\left(m^2+2m\right)=1>0\)

\(\Rightarrow\) Phương trình luôn có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=m+1-1=m\\x_2=m+1+1=m+2\end{matrix}\right.\)

\(\left|x_1\right|=3\left|x_2\right|\Leftrightarrow\left|m\right|=3\left|m+2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3m+6=-m\\3m+6=m\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=-\dfrac{3}{2}\\m=-3\end{matrix}\right.\)

a) Với m = 1 phương trình trở thành:

x 2  + 4x + 4 = 0 ⇔ (x + 2 ) 2  = 0 ⇔ x = -2

Vậy x = -2

b) Ta có: Δ' = m 2  - 5m + 4

Phương trình có hai nghiệm phân biệt

⇔ Δ' > 0 ⇔ m 2  - 5m + 4 > 0 

Do x1 < x2 < 1