2x(x - 1/7) = 0

<=> 2x = 0 hoặc x - 1/7 = 0

<=> x = 0 hoặc x = 1/7

phương trình <=>\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

vậy phương trình có hai nhiệm như trên đó

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-\frac{1}{7}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0:2\\x=0+\frac{1}{7}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)

2x(x-1/7)=0

=> 2x=0 hoặc x-1/7=0

=> x=0 hoặc x=1/7

vậy x=0 hoặc x=1/7

tk mk nha

a, Ta có 

\(x^2+2\left(2x+3\right)-\left|x^2-2x+2\right|=77\)

\(\Rightarrow x^2+4x+4+2-\left|x^2-2x+1+1\right|=77\)

\(\Rightarrow\left(x+2\right)^2+2-\left|\left(x-1\right)^2+1\right|=77\)

Vì \(\left(x+1\right)^2\ge0\left(\forall x\right)\) \(\Rightarrow\left(x+1\right)^2+1>0\Rightarrow\left|\left(x+1\right)^2+1\right|=\left(x+1\right)^2+1\)

\(\Rightarrow\left(x+2\right)^2-\left(x-1\right)^2+1=75\) \(\Rightarrow\left(x+2\right)^2-\left(x-1\right)^2=74\Rightarrow\left(x+2-x+1\right).\left(x+2+x+1\right)=74\)

\(\Rightarrow3.\left(x+3\right)=74\Rightarrow x+3=\frac{74}{3}\Rightarrow x=\frac{65}{3}\)

b, \(3x^2-2x+1=0\Rightarrow x^2-2x+1+2x^2=\left(x-1\right)^2+2x^2=0\)

Vì \(\left(x-1\right)^2\ge0;x^2\ge0\Rightarrow2x^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+2x^2\ge0\)

\(\Rightarrow\hept{\begin{cases}2x^2=0\\\left(x-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}}\) ( vô lý )

Vậy ko có gt x thỏa mãn

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

\(\Rightarrow x=0\) (hoặc) \(x-\frac{1}{7}=0\)

\(\Rightarrow\) x = 0 hoặc x = \(\frac{1}{7}\)

( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> ( 2x - 1 )² + ( x + 3 )² - 5( x² - 7² ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> x = -255/2

\(pt< =>4x^2-4x+1+x^2+6x+9-5x^2+5.49=0\)

\(< =>2x+255=0< =>x=-\frac{255}{2}\)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x+255=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=\frac{-255}{2}\)

Vậy tập nghiệm của phương trình là: \(S=\left\{\frac{-255}{2}\right\}\)

a , x.(2x+7)=0

(=) x = 0

     2x + 7 = 0 

(=) x = 0

     2x = -7

(=) x = 0

      x = -7/2

Mấy câu bạn hỏi có người hỏi rồi bạn tự tham khảo nhé

`x(2x+7)=0`

`<=>x=0` hoặc `2x+7=0`

`<=>x=0` hoặc `x=-7/2`

`x(2x+7)>0`

\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\2x+7>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\2x+7< 0\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>-\dfrac{7}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< -\dfrac{7}{2}\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x>0\\x< -\dfrac{7}{2}\end{matrix}\right.\)

`x(2x+7)<0`

\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\2x+7< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\2x+7>0\end{matrix}\right.\end{matrix}\right.\\ < =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< -\dfrac{7}{2}\end{matrix}\right.\left(voli\right)}\\\left\{{}\begin{matrix}x< 0\\x>-\dfrac{7}{2}\end{matrix}\right.\end{matrix}\right.\\ < =>-\dfrac{7}{2}< x< 0\)

2x(x - 1/7) = 0

=> 2x =  0  hoặc x - 1/7 = 0

=> x = 0 hoặc x = 1/7 

Đúng cho mình nha

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

\(\Leftrightarrow2x=0\) hoặc \(x-\frac{1}{7}=0\)

\(\Leftrightarrow x=0\) hoặc \(x=\frac{1}{7}\)