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:

Δ=(m+2)^2-4*2m=(m-2)^2

Để PT có hai nghiệm pb thì m-2<>0

=>m<>2

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1x_2}{4}\)

=>\(\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1x_2}{4}\)

=>\(\dfrac{m+2}{2m}=\dfrac{2m}{4}=\dfrac{m}{2}\)

=>2m^2=2m+4

=>m^2-m-2=0

=>m=2(loại) hoặc m=-1

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

\(\Delta=5^2-4\left(m-2\right)=25-4m+8=33-4m\)

Để pt có 2 nghiệm thì \(\Delta\ge0\Leftrightarrow33-4m\ge0\Leftrightarrow m\le\dfrac{33}{4}\)

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

\(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}=2\\ \Leftrightarrow\dfrac{x_2-1+x_1-1}{\left(x_1-1\right)\left(x_2-1\right)}=2\\ \Leftrightarrow x_1+x_2-2=2\left(x_1x_2-x_2-x_1+1\right)\\ \Leftrightarrow-5-2=2x_1x_2-2\left(x_1+x_2\right)+2\\ \Leftrightarrow2\left(m-2\right)-2.\left(-5\right)+2+7=0\\ \Leftrightarrow2m-4+10+2+7=0\\ \Leftrightarrow2m+15=0\\ \Leftrightarrow m=-\dfrac{15}{2}\left(tm\right)\)

\(x^2+5x+m-2\left(1\right)\)

PT (1) là PT bậc 2 có: \(\Delta=5^2-4.\left(m-2\right)=33-4m\)

Để PT có 2 nghiệm phân biệt \(x_1,x_2\) thì \(\Delta>0\Leftrightarrow33-4m>0\Leftrightarrow m< \dfrac{33}{4}\)

Theo định lý Viet ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{1}=-5\\x_1.x_2=\dfrac{c}{a}=\dfrac{m-2}{1}=m-2\end{matrix}\right.\)

Ta có: \(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}=2\Leftrightarrow\dfrac{x_2-1+x_1-1}{\left(x_1-1\right)\left(x_2-1\right)}=2\)

\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)-2}{x_1.x_2-\left(x_1+x_2\right)+1}=2\Leftrightarrow\dfrac{-5-2}{m-2-\left(-5\right)+1}=2\)

\(\Leftrightarrow\dfrac{-7}{m+4}=2\Leftrightarrow m+4=-\dfrac{7}{2}\Leftrightarrow m=-\dfrac{15}{2}\)

\(\Delta=25-4\left(m-2\right)=33-4m>0\Rightarrow m< \dfrac{33}{4}\)

Để biểu thức đề bài xác định \(\Rightarrow\) pt có nghiệm khác 1

\(\Rightarrow1+5+m-2\ne0\Rightarrow m\ne-5\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=m-2\end{matrix}\right.\)

\(\dfrac{1}{x_1-1}+\dfrac{1}{x_2-1}=2\Leftrightarrow\dfrac{x_1+x_2-2}{\left(x_1-1\right)\left(x_2-1\right)}=2\)

\(\Leftrightarrow\dfrac{x_1+x_2-2}{x_1x_2-\left(x_1+x_2\right)+1}=2\)

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

\(\Leftrightarrow-5-2=2\left(m-2\right)-2.\left(-5\right)+2\)

\(\Leftrightarrow m=-\dfrac{15}{2}\) (thỏa mãn)

Xét \(\Delta=4\left(m-1\right)^2-4.\left(-3\right)=4\left(m-1\right)^2+12>0\forall m\)

=>Pt luôn có hai nghiệm pb

Theo viet:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1.x_2=-3\ne0\forall m\end{matrix}\right.\)

Có \(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\)

\(\Leftrightarrow x_1^3+x_2^3=\left(m-1\right)x_1^2.x_2^2\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=\left(m-1\right).\left(-3\right)^2\)

\(\Leftrightarrow8\left(m-1\right)^3-3\left(-3\right).2\left(m-1\right)=9\left(m-1\right)\)

\(\Leftrightarrow8\left(m-1\right)^3+9\left(m-1\right)=0\)

\(\Leftrightarrow\left(m-1\right)\left[8\left(m-1\right)^2+9\right]=0\)

\(\Leftrightarrow m=1\)(do \(8\left(m-1\right)^2+9>0\) với mọi m)

Vậy m=1

Vì \(ac< 0\) \(\Rightarrow\) Phương trình luôn có 2 nghiệm phân biệt

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

Mặt khác: \(\dfrac{x_1}{x_2^2}+\dfrac{x_2}{x_1^2}=m-1\) \(\Rightarrow\dfrac{\left(x_1+x_2\right)\left(x_1^2+x_2^2-x_1x_2\right)}{x_1^2x_2^2}=m-1\)

  \(\Leftrightarrow\dfrac{\left(x_1+x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]}{x_1^2x_2^2}=m-1\)

  \(\Rightarrow\dfrac{\left(2m-2\right)\left(4m^2-8m+13\right)}{9}=m-1\)

  \(\Leftrightarrow...\)  

△'=(-2)²-1(m-1)

   =4-m+1

   =5-m

Để PT có 2 no pb thì △'>0

⇒5-m>0

⇒m<5

theo vi-ét ta có

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

mà: \(x^2_1x_2+x_1x_2^2-2\left(x_1+x_2\right)=0\)

⇔\(\left(x_1x_2\right)\left(x_1+x_2\right)-2\left(x_1+x_2\right)=0\)

⇔\(\left(m-1\right)4-2\cdot4=0\)

⇔\(4m-4-8=0\)

⇔4m-12=0

⇔4m=12

⇔m=3

Vậy ...

a. Với m=6 thì phương trình (1) có dạng 

x^2 - 5x +4= 0

<=> (x-1)(x-4)=0

<=> x=1 hoặc x=4

Vậy m=6 thì phương trình có nghiệm x=1 hoặc x=4

b. Xét \(\text{ Δ}=\left(-5\right)^2-4\cdot1\cdot\left(m-2\right)=33-4m\)

Để (1) có nghiệm phân biệt khi \(m< \dfrac{33}{4}\)

Theo Vi-et ta có: \(x_1x_2=m-2;x_1+x_2=5\)

Để 2 nghiệm phương trình (1) dương khi m>2

Ta có:

\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}+\dfrac{2}{\sqrt{x_1x_2}}=\dfrac{9}{4}\\ \Leftrightarrow\dfrac{5}{m-2}+\dfrac{2}{\sqrt{m-2}}=\dfrac{9}{4}\Leftrightarrow20+8\sqrt{m-2}=9\left(m-2\right)\\ \Leftrightarrow\left(\sqrt{m-2}-2\right)\left(9\sqrt{m-2}+10\right)=0\Leftrightarrow\sqrt{m-2}=2\Leftrightarrow m-2=4\Leftrightarrow m=6\left(t.m\right)\)

\(x^2-\left(m+1\right)x+m+4=0\left(1\right)\)

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

\(ddkt-thỏa:\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)

\(x1=0\Rightarrow\left(1\right)\Leftrightarrow m=-4\Rightarrow\left(1\right)\Leftrightarrow x^2+3x=0\Leftrightarrow\left[{}\begin{matrix}x1=0\\x2=-3< 0\left(loại\right)\end{matrix}\right.\)

\(x1\ne0\) \(\Rightarrow0< x1< x2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+1>0\\m+4>0\end{matrix}\right.\)\(\Rightarrow m>-1\)\(\left(3\right)\)

\(\left(2\right)\left(3\right)\Rightarrow m>5\)

\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\sqrt{3}\)

\(\Leftrightarrow x1+x2+2\sqrt{x1x2}=12\Leftrightarrow m+1+2\sqrt{m+4}=12\)

\(\Leftrightarrow m+4+2\sqrt{m+4}-15=0\)

\(đặt:\sqrt{m+4}=t>5\Rightarrow t^2+2t-15=0\Leftrightarrow\left[{}\begin{matrix}t=-5\left(ktm\right)\\t=3\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow m\in\phi\)

Để pt có 2 nghiệm pb 

\(\left(m+1\right)^2-4\left(m+4\right)=m^2+2m+1-4m-16\)

\(=m^2-2m-15>0\)

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

Ta có : \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=12\Leftrightarrow x_1+2\sqrt{x_1x_2}+x_2=12\)

Thay vào ta được \(m+1+2\sqrt{m+4}=12\Leftrightarrow2\sqrt{m+4}=11-m\)đk : m >= -4 

\(\Leftrightarrow4\left(m+4\right)=121-22m+m^2\Leftrightarrow m^2-26m+105=0\)

\(\Leftrightarrow m=21\left(ktm\right);m=5\left(ktm\right)\)