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

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

\(\rightarrow\)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

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

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

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

\(\sqrt{x^2+2x+1}-2=0\\ \Leftrightarrow\sqrt{\left(x+1\right)^2}=2\\ \Leftrightarrow\left|x+1\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

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

⇒ \(\left|x+1\right|=2\)

⇒ \(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

ko ai rảnh viết hết đâu bn

b, (x-2)(x+1)^2 + (x+1)(x-2)^2 = 0 

(x-2)(x+1)[(x+1)+(x-2)]=0 

(x-2)(x+1)(2x-1)=0 

Therefore, three possible answers for x: 

(2x-1) = 0, x = 1/2 

(x+1) = 0, x = -1 

(x-2) = 0, x = 2 

X = 2, -1 or 1/2 

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

\(a,Thaym=3.vào.\left(1\right),ta.được:x^2+5x+4=0\\ \Leftrightarrow x^2+x+4x+4=0\\ \Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+4\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\\ Vậy:S=\left\{-1;-4\right\}\\ b,\Delta=\left(m+2\right)^2-4.1.\left(m+1\right)=m^2+4m+4-4m-4=m^2\ge0\forall m\in R\\ \)

a. \(-x^2+14x-49=0\Leftrightarrow-\left(x-7\right)^2=0\)

\(\Leftrightarrow x-7=0\Leftrightarrow x=7\)

b. \(\left(x-1\right)^2+2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x-1+x+2\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow x=\dfrac{1}{2}\)

c. \(\left(2x-4\right)\left(3x+1\right)+\left(x-2\right)^2=0\)

\(\Leftrightarrow2\left(x-2\right)\left(3x+1\right)+\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\Leftrightarrow7x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

a)-x^2+14x-49=0

-(x^2-2.x.7+7^2)=0

-(x-7)^2=0

x-7=0

x=7

b)(x-1)^2+2(x-1)(x+2)+(x+2)^2=0

(x-1+x+2)^2=0

(2x+1)^2=0

2x+1=0

2x=-1

x=-1/2

c)(2x-4)(3x+1)+(x-2)^2=0

2(x-2)(3x+1)+(x-2)^2=0

(x-2)(2(3x+1)+x-2)=0

(x-2)7x=0

x-2=0 hoặc 7x=0

x=2 hoặc x=0

Bạn xem lại đề câu d nheNếu đúng đề chắc mình ko bit

\(a)x^3-\frac{x}{49}=0\)

\(\Leftrightarrow x\left(x^2-\frac{1}{7^2}\right)=0\)

\(\Leftrightarrow x=0\)Hoặc \(x^2-\frac{1}{7^2}=0\)

TH1: \(x\left(x^2-\frac{1}{7^2}\right)=0\\ x=\frac{0}{x^2-\frac{1}{7^2}}\\ \Leftrightarrow x=0\)

TH2: \(x\left(x^2-\frac{1}{7^2}\right)=0\\ x^2-\frac{1}{7^2}=\frac{0}{x}\\ x^2=0+\frac{1}{7^2}\\ x^2=\frac{1}{7^2}\\ x^2=\left(\frac{1}{7}\right)^2\\ \Leftrightarrow x=\frac{1}{7}\)

Vậy \(x=0\)Hoặc \(x=\frac{1}{7}\)

a) x³ - x/49 = 0

<=> x(x² - 1/49) = 0

<=> x = 0 hoặc x² - 1/49 = 0

<=> x = 0 hoặc x = +1/7

b) x² - 7x + 12 = 0

<=> (x - 3)(x - 4) = 0

<=> x - 3 = 0 hoặc x - 4 = 0

<=> x = 3 hoặc x = 4

c) 4x² - 3x - 1 = 0

<=> 4x² + x - 4x - 1 = 0

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

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

<=> 4x + 1 = 0 hoặc x - 1 = 0

<=> x = -1/4 hoặc x = 1

d) x³ - 2x - 4 = 0

<=> (x² + 2x + 2)(x - 2) = 0

vì x² + 2x + 2 khác 0 nên:

<=> x - 2 = 0

<=> x = 2