\(\Delta'=4-\left(-2m+1\right)=2m+5\)

Để pt có 2 nghiệm x1;x2 

\(2m+5\ge0\Leftrightarrow m\ge-\dfrac{5}{2}\)

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-4\left(1\right)\\x_1x_2=-2m+1\left(2\right)\end{matrix}\right.\)

Lại có \(x_1-3x_2=12\left(3\right)\)

Từ (1) ; (3) ta có \(\left\{{}\begin{matrix}x_1+x_2=-4\\x_1-3x_2=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x_2=-16\\x_1=-4-x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=-4\\x_1=0\end{matrix}\right.\)

Thay vào (3) ta đc 

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

a, Với m = -2 pt có dạng 

\(x^2+4x-5=0\)

ta có : a + b + c = 1 + 4 - 5 = 0 

nên pt có 2 nghiệm \(x=1;x=-5\)

b, delta' = m^2 - ( m^2 - 9 ) = 9 > 0 

Vậy pt luôn có 2 nghiệm pb 

Theo Vi et : x1 + x2 = 2m ; x1x2 = m^2 - 9

Ta có : x1^2 + x2^2(x1+x2) = 12

<=> x1^2 + 2x2^2m = 12 

đề có thiếu dấu ko bạn ? 

a: Thay m=-2 vào pt, ta được:

\(x^2-2\cdot\left(-2\right)\cdot x+\left(-2\right)^2-9=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)

=>x=-5 hoặc x=1

a: \(\left\{{}\begin{matrix}x_1+x_2=-b\\x_1x_2=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=10\\c=-24\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}x_1+x_2=-b\\x_1x_2=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-b=-5\\c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=5\\c=0\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=1-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-2\\c=-1\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}x_1+x_2=3-\dfrac{1}{2}=\dfrac{5}{2}\\x_1x_2=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{5}{2}\\c=-\dfrac{3}{2}\end{matrix}\right.\)

\(ac=-12< 0\) nên pt luôn có 2 nghiệm pb trái dấu

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=-12\end{matrix}\right.\)

\(x_1^2-x_2^2-14\left(m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)-14\left(m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(m+1\right)-14\left(m+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}m=-1\\x_1-x_2=7\left(1\right)\end{matrix}\right.\)

Xét (1), kết hợp với Viet ta được: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1-x_2=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2m+9}{2}\\x_2=\dfrac{2m-5}{2}\end{matrix}\right.\)

Thế vào \(x_1x_2=-12\Leftrightarrow\left(\dfrac{2m+9}{2}\right)\left(\dfrac{2m-5}{2}\right)=-12\)

\(\Leftrightarrow4m^2+8m+3=0\Rightarrow\left[{}\begin{matrix}m=-\dfrac{3}{2}\\m=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(m=\left\{-1;-\dfrac{3}{2};-\dfrac{1}{2}\right\}\)

\(x^2-2x-\sqrt{3}+1=0\)

\(\Delta=b^2-4ac=4-4\left(-\sqrt{3}+1\right)=4\sqrt{3}>0\)

\(\rightarrow\)Phương trình có 2 nghiệm phân biệt

Theo vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=2\\P=x_1x_2=\dfrac{c}{a}=-\sqrt{3}+1\end{matrix}\right.\)

\(M=x_1^2x_2^2-2x_1x_2-x_1-x_2\)

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

\(=\left(-\sqrt{3}+1\right)^2-2\left(-\sqrt{3}+1\right)-2\)

\(=0\)

\(\Delta=1-m+3=4-m\)

Để pt có ng₀ thì \(\Delta\ge0\)\(\Rightarrow m\le4\)

Theo hệ thức Viet: \(x_1+x_2=2;x_1x_2=m-3\)

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

\(\Leftrightarrow x_1-x_2=6\)

Có hpt:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1-x_2=6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x_1=4\\x_2=-2\end{matrix}\right.\)

\(\Rightarrow x_1x_2=-8=m-3\Leftrightarrow m=-5\)(TM)

Câu a thì cứ thay m = 1 vào rồi giải nha :)

\(b.pt:x^2-2mx+4m-4=0\)

\(\Delta=\left(-2m\right)^2-4.\left(4m-4\right)=\left(2m-4\right)^2\ge0\forall m\left(1\right)\)

Để pt có 2 nghiệm pb thì \(\Delta>0\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow2m-4\ne0\Leftrightarrow m\ne2\)

Theo hệ thức Vi-et:

\(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=4m-4\end{matrix}\right.\)

Thay nghiệm x₁ vào pt ban đầu: \(x^2_1-2mx_1+4m-4=0\Leftrightarrow x^2_1=2mx_1-4m+4\left('\right)\)

Thay \(\left('\right)\) vào điều kiện thõa mãn: \(2mx_1-4m+4+\left(x_1+x_2\right)x_2=12\)

\(\Leftrightarrow2mx_1-4m+4+2mx_2=12\Leftrightarrow2m\left(x_1+x_2\right)-4m+4=12\Leftrightarrow4m^2-4m+4=12\Leftrightarrow4m^2-4m-8=0\)

\(\Delta=\left(-4\right)^2-4.4.\left(-8\right)=144>0\)

\(\Rightarrow\left\{{}\begin{matrix}m_1=\frac{4+\sqrt{144}}{2.4}=2\left(KTM\right)\\m_2=\frac{4-\sqrt{144}}{2.4}=-1\end{matrix}\right.\)

Vậy \(m=-1\)