bạn đăng tách ra cho mn giúp nhé 

a, Để pt có 2 nghiệm pb 

\(\Delta'=1-m\ge0\Leftrightarrow m\le1\)

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-2\left(1\right)\\x_1x_2=m\left(2\right)\end{matrix}\right.\)

\(x_1-3x_2=0\)(3) 

Từ (1) ; (3) ta có hệ \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1-3x_2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x_1=-2\\x_2=-2-x_1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-\dfrac{1}{2}\\x_2=-\dfrac{3}{2}\end{matrix}\right.\)

Thay vào (2) ta được \(m=\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)=\dfrac{3}{4}\)

\(b,\Delta=\left(m+5\right)^2-4\left(-m+6\right)\ge0\Leftrightarrow\left[{}\begin{matrix}m\le-7-4\sqrt{3}\\m\ge-7+4\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=m+5\\2x1+3x2=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x1+2x2=2m+10\\2x1+3x2=13\end{matrix}\right.\)\(\)

\(\Rightarrow x2=13-2m-10=3-2m\Rightarrow x1=m+5-x2=m+5-3+2m=3m+2\)

\(x1x2=6-m\Rightarrow\left(3-2m\right)\left(3m+2\right)=6-m\Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=1\left(tm\right)\end{matrix}\right.\)

\(c,\Delta'=\left(m+1\right)^2-\left(m^2-2m+29\right)\ge0\Leftrightarrow m\ge7\)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1=2x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x2=\dfrac{2m+2}{3}\\x1=\dfrac{2\left(2m+2\right)}{3}\end{matrix}\right.\)

\(\Rightarrow x1.x2=\dfrac{\left(2m+2\right).2\left(2m+2\right)}{9}=m^2-2m+29\Leftrightarrow\left[{}\begin{matrix}m=11\left(tm\right)\\m=23\left(tm\right)\end{matrix}\right.\)

Đáp án A

Phương trình đã cho có nghiệm phân biệt khi : 

\(\Delta'=m^2-\left(m^2+2m+3\right)=-2m-3>0\)

\(\Leftrightarrow m< -\dfrac{3}{2}\)(*)

Hệ thức Viette : \(\left\{{}\begin{matrix}x_1+x_2=-2m\\x_1x_2=m^2+2m+3\end{matrix}\right.\)

Có \(x_1^3+x_2^3=108\)

\(\Leftrightarrow\left(x_1+x_2\right).\left(x_1^2-x_1x_2+x_2^2\right)=108\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=108\)

\(\Leftrightarrow-8m^3+6m\left(m^2+2m+3\right)=108\)

\(\Leftrightarrow m^3-6m^2-9m+54=0\)

\(\Leftrightarrow\left(m-6\right).\left(m-3\right).\left(m+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=6\\m=\pm3\end{matrix}\right.\)

Kết hợp (*) được m = -3 thỏa mãn

Tham khảo:

\(\Delta'=1-\left(m-3\right)=4-m>0\Rightarrow m< 4\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-3\end{matrix}\right.\)

Do \(x_1+x_2=2\Rightarrow x_2=2-x_1\)

Ta có:

\(x_1^2+x_1x_2=2x_2-12\)

\(\Leftrightarrow x_1\left(x_1+x_2\right)=2\left(2-x_1\right)-12\)

\(\Leftrightarrow2x_1=4-2x_1-12\)

\(\Leftrightarrow4x_1=-8\Rightarrow x_1=-2\Rightarrow x_2=4\)

Thế vào \(x_1x_2=m-3\Rightarrow m-3=-8\)

\(\Rightarrow m=-5\)

Δ=(-2)^2-4(m-1)=4-4m+4=8-4m

Để phương trình có hai nghiệm thì 8-4m>=0

=>m<=2

x1+x2=2; x1x2=m-1

=>x1=2-x2

=>x1+1=3-x2

x1^2+x2^2=(x1+x2)^2-2x1x2=2^2-2(m-1)=4-2m+2=6-2m

=>x1^2=6-2m-x2^2

2x1(x1-x2)+3=7m+(x2+2)^2

=>2x1^2-2x1x2+3=7m+x2^2+2x2+4

=>2(6-2m-x2^2)-2x1x2+3-7m-x2^2-2x2-4=0

=>2(6-2m-x2^2)-2x2(3-x2)-7m-1=0

=>12-4m-2x2^2-6x2-2x2^2-7m-1=0

=>-4x2^2-6x2-11m+11=0

=>4x2^2+6x2+11m-11=0(1)

Để phương trình (1) có nghiệm thì 6^2-4*4*(11m-11)>=0

=>36-16(11m-11)>=0

=>16(11m-11)<=36

=>11m-11<=9/4

=>11m<=53/4

=>m<=53/44