\(\Delta=\left(2m-3\right)^2-4\left(2m-4\right)=\left(2m-5\right)^2\ge0;\forall m\)

Pt luôn có 2 nghiệm với mọi m

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m-3\\x_1x_2=2m-4\end{matrix}\right.\)

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{2}\)

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

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

\(\Rightarrow4m-6=2m-4\)

\(\Leftrightarrow2m=2\)

\(\Leftrightarrow m=1\) (thỏa mãn)

\(x^2+6x+2m-3=0\)

\(\Delta=6^2-4\cdot1\cdot\left(2m-3\right)\)

\(=36-8m+12=-8m+48\)

Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)

=>-8m+48>0

=>-8m>-48

=>m<6

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-6\\x_1x_2=\dfrac{c}{a}=2m-3\end{matrix}\right.\)

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

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

=>\(\dfrac{-6-2}{x_1x_2-\left(x_1+x_2\right)+1}=-6+2=-4\)

=>\(x_1x_2-\left(x_1+x_2\right)+1=\dfrac{-8}{-4}=2\)

=>2m-3-(-6)=2

=>2m-3+6=2

=>2m+3=2

=>2m=-1

=>\(m=-\dfrac{1}{2}\left(nhận\right)\)

ĐK:`x_1,x_2 ne 0=>x_1.x_2 ne 0`

`=>-2m-1 ne 0=>m ne -1/2`

Ta có:`a=1,b=2m,c=-2m-1`

`=>a+b+c=1+2m-2m-1=0`

`<=>` \(\left[ \begin{array}{l}x=1\\x=-2m-1\end{array} \right.\) 

PT có 2 nghiệm pn

`=>-2m-1 ne 1`

`=>-2m ne 2`

`=>m ne -1`

Nếu `x_1=1,x_2=-2m-1`

`pt<=>6=1+1/(-2m-1)`

`<=>5=1/(-2m-1)`

`<=>2m+1=-1/5`

`<=>2m=-6/5`

`<=>m=-3/5(tm)`

Nếu `x_2=1,x_1=-2m-1`

`pt<=>6/(-2m-1)=-2m-1+1=-2m`

`<=>6/(2m+1)=2m`

`<=>3/(2m+1)=m`

`<=>2m^2+m-3=0`

`a+b+c=0`

`=>m_1=1(tm),m_2=-c/a=-3/2(tm)`

Vậy `m in {-3/5,1,-3/2}` thì ....

a: \(\text{Δ}=\left(2m+1\right)^2-4m\left(m+3\right)\)

\(=4m^2+4m+1-4m^2-12m\)

\(=-8m+1\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

\(\Leftrightarrow-8m+1>0\)

\(\Leftrightarrow-8m>-1\)

hay \(m< \dfrac{1}{8}\)

`1)`

$a\big)\Delta=7^2-5.4.1=29>0\to$ PT có 2 nghiệm pb

$b\big)$

Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{7}{5}\\x_1x_2=\dfrac{1}{5}\end{matrix}\right.\)

\(A=\left(x_1-\dfrac{7}{5}\right)x_1+\dfrac{1}{25x_2^2}+x_2^2\\ \Rightarrow A=\left(x_1-x_1-x_2\right)x_1+\left(\dfrac{1}{5}\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\\ \Rightarrow A=-x_1x_2+\left(x_1x_2\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\)

\(\Rightarrow A=-x_1x_2+x_1^2+x_2^2\\ \Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2\\ \Rightarrow A=\left(\dfrac{7}{5}\right)^2-3\cdot\dfrac{1}{5}=\dfrac{34}{25}\)

a) Với m = 2, phương trình đã cho trở thành:

2x² - 6x + 2.2 - 5 = 0

⇔ 2x² - 6x - 1 = 0

∆' = (-3)² - 2.(-1) = 11 > 0

⇒ Phương trình có 2 nghiệm phân biệt:

x₁ = [-(-3) + 11]/2 = (3 + 11)/2

x₂ = [-(-3) - 11]/2 = (3 - 11)/2

b) ∆' = (-3)² - 2.(2m - 5)

= 9 - 4m + 10

= 19 - 4m

Để phương trình đã cho có nghiệm thì ∆' ≥ 0

⇔ 19 - 4m ≥ 0

⇔ 4m ≤ 19

⇔ m ≤ 19/4

Theo định lý Viét, ta có:

x₁ + x₂ = 3

x₁x₂ = (2m - 5)/2

Ta có:

1/x₁ + 1/x₂ = 6

⇔ (x₁ + x₂)/(x₁x₂) = 6

⇔ 3/[(2m - 5)/2] = 6

⇔ (2m - 5)/2 = 1/2

⇔ 2m - 5 = 1

⇔ 2m = 6

⇔ m = 3 (nhận)

Vậy m = 3 thì phương trình đã cho có 2 nghiệm thỏa mãn yêu cầu

Đề sai rồi bạn

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

Để pt vô nghiệm thì \(m^2-4< 0\Leftrightarrow-2< m< 2\)

Để pt có nghiệm kép thì \(m^2-4=0\Leftrightarrow m=\pm2\)

Để pt có 2 nghiệm phân biệt thì \(m^2-4>0\Leftrightarrow\left[{}\begin{matrix}m< -2\\m>2\end{matrix}\right.\)

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

\(a,ĐKXĐ:x_1,x_2\ne0\\ \dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=2\\ \Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=2\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\\ \Leftrightarrow\left(2m-2\right)^2-4\left(-2m+5\right)=0\\ \Leftrightarrow4m^2-8m+4+8m-20=0\\ \Leftrightarrow4m^2-16=0\\ \Leftrightarrow m=\pm2\)

\(b,x_1+x_2+2x_1x_2\le6\\ \Leftrightarrow2m-2+2\left(-2m+5\right)\le6\\ \Leftrightarrow2m-2-4m+10-6\le0\\ \Leftrightarrow-2m+2\le0\\ \Leftrightarrow m\ge1\)

1. Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\\x_1.x_2=\dfrac{1}{3}\end{matrix}\right.\)

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

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

  \(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}=\dfrac{\dfrac{22}{9}}{\dfrac{8}{3}}=\dfrac{11}{12}\)

\(1,3x^2+4x+1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{4}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)

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

\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}\)

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

\(=\dfrac{S^2-2P-S}{P-S+1}\)

\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}\)

\(=\dfrac{11}{12}\)

Vậy \(C=\dfrac{11}{12}\)

\(3,3x^2-7x-1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{7}{3}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1\)

\(=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)

\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(x_1^2+x_2^2\right)+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(S^2-2P\right)+2P}{S}\)

\(=\dfrac{2\left(\dfrac{7}{3}^2-2\left(-\dfrac{1}{3}\right)\right)+2\left(-\dfrac{1}{3}\right)}{\dfrac{7}{3}}\)

\(=\dfrac{104}{21}\)

Vậy \(B=\dfrac{104}{21}\)