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 \(m=\left(-\dfrac{1}{2}\right)\left(-\dfrac{3}{2}\right)=\dfrac{3}{4}\)

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

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

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

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

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

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

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

Ta có: \(\Delta=4m^2-8m+1\)

Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow\Delta>0\) \(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{2-\sqrt{3}}{2}\\x>\dfrac{2+\sqrt{3}}{2}\end{matrix}\right.\)

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=1-2m\left(1\right)\\x_1x_2=m\left(2\right)\end{matrix}\right.\)

Ta lập được HPT \(\left\{{}\begin{matrix}x_1+x_2=1-2m\\2x_1=x_2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x_1=1-2m\\x_2=2x_1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{1-2m}{3}\\x_2=\dfrac{2-4m}{3}\end{matrix}\right.\)

Kết hợp với (2), ta được:

\(\dfrac{8m^2-12m+2}{9}=m\) \(\Leftrightarrow...\) 

Lời giải:

Để pt có 2 nghiệm thì:

$\Delta'=(m-1)^2+2m-5\geq 0$

$\Leftrightarrow m^2-4\geq 0$

$\Leftrightarrow m\geq 2$ hoặc $m\leq -2$
Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2(1-m)\\ x_1x_2=-2m+5\end{matrix}\right.\)

\(2x_1+3x_2=-5\)

\(\Leftrightarrow 2(x_1+x_2)+x_2=-5\Leftrightarrow 4(1-m)+x_2=-5\)

\(\Leftrightarrow x_2=4m-9\)

\(x_1=2(1-m)-x_2=11-6m\)

$x_1x_2=-2m+5$

$\Leftrightarrow (4m-9)(11-6m)=-2m+5$

Giải pt này suy ra $m=2$ hoặc $m=\frac{13}{6}$ (đều thỏa mãn)

Lời giải:
Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=(m-1)^2+2m+1=m^2+2\geq 0$

$\Leftrightarrow m\in\mathbb{R}$
Áp dụng định lý Viet:

$x_1+x_2=2(m-1)$

$x_1x_2=-2m-1$

Khi đó:

$2x_1+3x_2+3x_1x_2=-11$

$\Leftrightarrow 2(x_1+x_2)+3x_1x_2+x_2=-11$

$\Leftrightarrow 4(m-1)+3(-2m-1)+x_2=-11$

$\Leftrightarrow x_2=2m-4$

$x_1=2(m-1)-x_2=2m-2-(2m-4)=2$

$-2m-1=x_1x_2=2(2m-4)$

$\Leftrightarrow -2m-1=4m-8$

$\Leftrightarrow 7=6m$

$\Leftrightarrow m=\frac{7}{6}$

Lời giải:
Để pt có 2 nghiệm $x_1,x_2$ thì:

$\Delta'=(m-1)^2+2m+1=m^2+2\geq 0$

$\Leftrightarrow m\in\mathbb{R}$
Áp dụng định lý Viet:

$x_1+x_2=2(m-1)$

$x_1x_2=-2m-1$

Khi đó:

$2x_1+3x_2+3x_1x_2=-11$

$\Leftrightarrow 2(x_1+x_2)+3x_1x_2+x_2=-11$

$\Leftrightarrow 4(m-1)+3(-2m-1)+x_2=-11$

$\Leftrightarrow x_2=2m-4$

$x_1=2(m-1)-x_2=2m-2-(2m-4)=2$

$-2m-1=x_1x_2=2(2m-4)$

$\Leftrightarrow -2m-1=4m-8$

$\Leftrightarrow 7=6m$

$\Leftrightarrow m=\frac{7}{6}$

\(ac=-2< 0\Rightarrow\) phương trình luôn có 2 nghiệm pb trái dấu

Mà \(x_1>x_2\Rightarrow\left\{{}\begin{matrix}x_2< 0\\x_1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x_1\right|=x_1\\\left|x_2\right|=-x_2\end{matrix}\right.\)

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

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

\(\Leftrightarrow2x_1+x_2=2+x_1\)

\(\Leftrightarrow x_1+x_2=2\)

\(\Leftrightarrow m-1=2\)

\(\Rightarrow m=3\)

b) phương trình có 2 nghiệm  \(\Leftrightarrow\Delta'\ge0\)

\(\Leftrightarrow\left(m-1\right)^2-\left(m-1\right)\left(m+3\right)\ge0\)

\(\Leftrightarrow m^2-2m+1-m^2-3m+m+3\ge0\)

\(\Leftrightarrow-4m+4\ge0\)

\(\Leftrightarrow m\le1\)

Ta có: \(x_1^2+x_1x_2+x_2^2=1\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=1\)

Theo viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)\\x_1x_2=\dfrac{c}{a}=m+3\end{matrix}\right.\)

\(\Leftrightarrow\left[-2\left(m-1\right)^2\right]-2\left(m+3\right)=1\)

\(\Leftrightarrow4m^2-8m+4-2m-6-1=0\)

\(\Leftrightarrow4m^2-10m-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m_1=\dfrac{5+\sqrt{37}}{4}\left(ktm\right)\\m_2=\dfrac{5-\sqrt{37}}{4}\left(tm\right)\end{matrix}\right.\Rightarrow m=\dfrac{5-\sqrt{37}}{4}\)

Câu c:

Ta có: \(\Delta=4m^2+4m-11\)

Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow4m^2+4m-11>0\)

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=2m+5\end{matrix}\right.\)

Để phương trình có 2 nghiệm dương phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+4m-11>0\\2m+3>0\\2m+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< \dfrac{-1-2\sqrt{3}}{2}\\m>\dfrac{-1+2\sqrt{3}}{2}\end{matrix}\right.\\m>-\dfrac{3}{2}\\m>-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow m>\dfrac{-1+2\sqrt{3}}{2}\)

 Mặt khác: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\)

\(\Rightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{16}{9}\) \(\Rightarrow\dfrac{2m+3+2\sqrt{2m+5}}{2m+5}=\dfrac{16}{9}\)

\(\Rightarrow18m+27+18\sqrt{2m+5}=32m+80\)

\(\Leftrightarrow14m-53=18\sqrt{2m+5}\)

\(\Rightarrow\) ...

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