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

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

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

Theo vi-ét ta có :

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

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

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

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

\(=0\)

b) 5x(x-2000)-x+2000=0

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ai giúp minh làm bài 5 phía trên với

c) Ta có: \(2x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)

d) Ta có: \(5x^2+x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

a: Khi m=1 thì (1) sẽ là:

x^2-x-8=0

=>\(x=\dfrac{1\pm\sqrt{33}}{2}\)

b: 3x1^2+3x2^2+2x1x2=5

=>3[(x1+x2)^2-2x1x2]+2x1x2=5

=>3[(2m-1)^2-2(-8m)]+2(-8m)=5

=>3(4m^2-4m+1+16m)-16m=5

=>12m^2+36m+3-16m-5=0

=>12m^2+20m-2=0

=>\(m=\dfrac{-5\pm\sqrt{31}}{6}\)

a: Khi x=2 thì pt sẽ là 2^2-2(m-1)*2-2m-1=0

=>4-2m-1-4(m-1)=0

=>-2m+3-4m+4=0

=>-6m+7=0
=>m=7/6

vậy  2x³-x²+5x+3 = 0  <=> (2x+1).(x² - x+ 3) = 0 <=> 2x+1 = 0 hoặc x² - x + 3 = 0

+) 2x+1 = 0 <=> x = -1/2

+) x² - x+ 3 = 0 Vô nghiệm vì x² - x+ 3  = x² - 2.x. \(\frac{1}{2}\)+ \(\frac{1}{4}\) + \(\frac{11}{4}\) = (x -  \(\frac{1}{2}\)) ² + \(\frac{11}{4}\)> 0 với mọi x

Vậy phương trình có nghiệm x = -1/2

Đk: 3m - 1 >= 0 <=> m>= 1/3

Để phương trình có nghiệm kép 

<=> \(\Delta=4.\left(3m-1\right)-4\sqrt{m^2-6m+17}=0\)

<=> 9m² - 6m + 1 = m² - 6m + 17

<=> 8m² = 16

<=> \(m=\sqrt{2}\)(Vì m >= 1/3).

Vậy với m = căn 2 thì phương trình có nghiệm kép.

x1 = x2 = \(-2\sqrt{3\sqrt{2}-1}\)

ẩn x tham số m

\(\Leftrightarrow\left[\left(m^2-m-1\right)-\left(3-m\right)\right]x>5m\)

\(\Leftrightarrow\left(m^2-4\right)x>5m\)

\(m=2;\Leftrightarrow0.x>5.2=>vo.N_0\)

\(m=-2\Leftrightarrow0.x>-10;N_0\forall\in R\)

\(\left|m\right|< 2\Leftrightarrow x< \dfrac{5m}{m^2-4}\)

\(\left|m\right|>2\Leftrightarrow x>\dfrac{5m}{m^2-4}\)

a) \(\dfrac{2x+1}{x-2}=3\Rightarrow2x+1=3x-6\Rightarrow x=7\)

b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\Rightarrow4x-6=x+1\Rightarrow3x=7\Rightarrow x=\dfrac{7}{3}\)

a) \(\dfrac{2x+1}{x-2}=3\)

dkxd : x ≠ 2

MTC : x - 2

Quy đồng mẫu thức : 

⇒ \(\dfrac{2x+1}{x-2}=\dfrac{3\left(x-2\right)}{x-2}\)

Suy ra : 2x + 1 = 3(x - 2)

      \(\) \(\Leftrightarrow\) 2x + 1 = 3x - 6

       \(\Leftrightarrow\) 2x + 1 - 3x + 6 = 0

      \(\Leftrightarrow\) -1x + 7 = 0

     \(\Leftrightarrow\) -1x = -7

    \(\Leftrightarrow\) x = \(\dfrac{-7}{-1}=7\)

 Vậy S = \(\left\{7\right\}\)

b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\)

dkxd : x ≠ -1

MTC : 2(x + 1)

Quy đồng mẫu thức : 

⇒ \(\dfrac{2\left(2x-3\right)}{2\left(x+1\right)}=\dfrac{1\left(x+1\right)}{2\left(x+1\right)}\)

Suy ra : 2(2x - 3) = x + 1

        \(\Leftrightarrow\) 4x - 6 - x - 1 = 0

       \(\Leftrightarrow\) 3x - 7 = 0

       \(\Leftrightarrow\) 3x = 7

      \(\Leftrightarrow\) x = \(\dfrac{7}{3}\)

   Vậy S = \(\left\{\dfrac{7}{3}\right\}\)

 Chúc bạn học tốt

a: Ta có: \(\dfrac{2x+1}{x-2}=3\)

\(\Leftrightarrow3x-6=2x+1\)

hay x=7

b: Ta có: \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\)

\(\Leftrightarrow4x-6=x+1\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)