ĐKXĐ: m<>-1

Ta có: \(\Delta=\left[-2\left(m-1\right)\right]^2-4\left(m+1\right)\left(m-2\right)\)

\(=\left(2m-2\right)^2-4\left(m^2-m-2\right)\)

\(=4m^2-8m+4-4m^2+4m-8\)

\(=-4m-4\)

Để phương trình có hai nghiệm phân biệt thì -4m-4>0

hay m<-1

Áp dụng hệ thức Vi-et, ta được:

\(\left\{{}\begin{matrix}x_1\cdot x_2=\dfrac{m-2}{m+1}\\x_1+x_2=\dfrac{2\left(m-1\right)}{m+1}\end{matrix}\right.\)

\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=4\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=4x_1x_2\)

\(\Leftrightarrow\left(\dfrac{2m-2}{m+1}\right)^2-6\cdot\dfrac{m-2}{m+1}=0\)

\(\Leftrightarrow\left(2m-2\right)^2-6\left(m^2-m-2\right)=0\)

\(\Leftrightarrow4m^2-8m+4-6m^2+6m+12=0\)

\(\Leftrightarrow-2m^2-2m+16=0\)

\(\Leftrightarrow m^2-m-8=0\)

Đến đây bạn tự giải nhé

PT có 2 nghiệm \(\Leftrightarrow\Delta=4\left(m-1\right)^2-4\left(m-2\right)\left(m+1\right)\ge0\)

\(\Leftrightarrow4m^2-8m+4-4m^2+4m+8\ge0\\ \Leftrightarrow12-4m\ge0\\ \Leftrightarrow m\le3\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{m+1}\\x_1x_2=\dfrac{m-2}{m+1}\end{matrix}\right.\)

\(\dfrac{x_2}{x_1}+\dfrac{x_1}{x_2}=-4\\ \Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=-4\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=-4x_1x_2\\ \Leftrightarrow\left(x_1+x_2\right)^2=-2x_1x_2\\ \Leftrightarrow\dfrac{4\left(m-1\right)^2}{\left(m+1\right)^2}=\dfrac{4-2m}{m+1}\\ \Leftrightarrow4\left(m-1\right)^2=\left(4-2m\right)^2\\ \Leftrightarrow4m^2-8m+4=16-16m+4m^2\\ \Leftrightarrow8m=12\Leftrightarrow m=\dfrac{3}{2}\left(tm\right)\)

Hai vecto đã cho cùng phương khi:

\(\dfrac{m}{4}=\dfrac{1}{-2}\Rightarrow m=-2\)

Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(7;-4\right)\\\overrightarrow{DC}=\left(3-x;7-y\right)\end{matrix}\right.\)

ABCD là hbh khi: \(\overrightarrow{AB}=\overrightarrow{DC}\)

\(\Rightarrow\left\{{}\begin{matrix}3-x=7\\7-y=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-4\\y=11\end{matrix}\right.\)

\(\Rightarrow D\left(-4;11\right)\)

2x + y = 1 <=> y = 1 - 2x

Thế vào pt còn lại thì:

x^2 + (1 - 2x)^2 - x(1 - 2x) = 3

<=> x^2 + 4x^2 - 4x + 1 - x + 2x^2 - 3 = 0

<=> 7x^2 - 5x - 2 = 0

<=> (x - 1)(7x + 2) = 0

<=> x = 1 hoặc x = -2/7

Với x = 1 <=> y = 1 - 2.1 = -1

Với x = -2/7 <=> y = 1 - 2.(-2/7) = 11/7

\(ĐK:x\le3\\ x+\sqrt{2x^2-3x-1}=3\\ \Leftrightarrow\sqrt{2x^2-3x-1}=3-x\\ \Leftrightarrow2x^2-3x-1=x^2-6x+8\\ \Leftrightarrow x^2+3x-9=0\\ \Delta=9+36=45\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3-3\sqrt{5}}{2}\left(ktm\right)\\x=\dfrac{-3+3\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)

D

Pt đã cho có 2 nghiệm trái dấu khi:

\(ac< 0\Leftrightarrow\left(m+1\right)\left(m-2\right)< 0\)

\(\Leftrightarrow-1< m< 2\)

A

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-1=0\\m^2-2m-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=\pm1\\m\ne-1;m\ne3\end{matrix}\right.\Leftrightarrow m=1\)

Chọn A