Pt có 2 nghiệm khi: \(\left\{{}\begin{matrix}m\ne0\\\Delta'=9\left(m-1\right)^2-9m\left(m-3\right)\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ge-1\end{matrix}\right.\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{6\left(m-1\right)}{m}\\x_1x_2=\dfrac{9\left(m-3\right)}{m}\end{matrix}\right.\)

\(x_1+x_2=x_1x_2\Rightarrow\dfrac{6\left(m-1\right)}{m}=\dfrac{9\left(m-3\right)}{m}\)

\(\Rightarrow6\left(m-1\right)=9\left(m-3\right)\)

\(\Rightarrow m=7\)

A đúng

Chọn 

6.B

Hàm nghịch biến trên R khi:

\(1-m< 0\Rightarrow m>1\)

5.B

Đồ thị đi qua A nên:

\(-1=2a-2\Rightarrow2a=1\Rightarrow a=\dfrac{1}{2}\)

TH1: thể 3 ở A, B, E

A-B-ddE- = 3.3.2.1.2 = 36

TH2: Thể 3 ở d

A-B-ddE- = 2.2.1.2 = 8

Tổng là 44

\(\Leftrightarrow\dfrac{3\left(2x-1\right)-36}{12}=\dfrac{2\left(1+x\right)}{12}\)

\(\Leftrightarrow\dfrac{6x-3-36}{12}=\dfrac{2+2x}{12}\)

\(\Leftrightarrow6x-39=2+2x\)

\(\Leftrightarrow4x=41\Leftrightarrow x=\dfrac{41}{4}\)

\(\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow4-4x+x^2=4x^2-1\\ \Leftrightarrow4x^2-1-x^2+4x-4=0\\ \Leftrightarrow3x^2+4x-5=0\)

Đến đây mik thấy nghiệm rất xấu bạn xem đề đúng chx nhé

\(3.\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\)

\(\Leftrightarrow3.\left(2-x\right)=2x+1\)

\(\Leftrightarrow6-3x=2x+1\)

\(\Leftrightarrow-3x-2x=1-6\)

\(\Leftrightarrow-5x=-5\)

\(\Leftrightarrow x=1\)

Vậy: S = {1}

\(3\left(2-x\right)^2=\left(2x-1\right)\left(2x+1\right)\\ \Leftrightarrow3\left(4-4x+x^2\right)=4x^2-1\\ \Leftrightarrow12-12x+3x^2=4x^2-1\\ \Leftrightarrow4x^2-1-3x^2+12x-12=0\\ \Leftrightarrow x^2+12x-13=0\\ \Leftrightarrow x^2+13x-x-13=0\\ \Leftrightarrow x\left(x+13\right)-\left(x+13\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-13\end{matrix}\right.\)

\(\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \left(x-7\right)\left(2x-5-5-x\right)=0\\ \left(x-7\right)\left(x-10\right)=0\\ \left\{{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\left\{{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)

\(\left(2x-5\right)\left(x-7\right)=\left(x-7\right)\left(5+x\right)\\ \Leftrightarrow\left(2x-5\right)\left(x-7\right)-\left(x-7\right)\left(5+x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(2x-5-5-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=10\end{matrix}\right.\)