8x(x-2017)-2x+4034=0

\(\Leftrightarrow\)8x(x-2017)-2(x-2017)=0

\(\Leftrightarrow\)(x-2017)(8x-2)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2017=0\\8x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2017\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy x\(\in\left\{2017;\frac{1}{4}\right\}\)

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(8x\left(x-2017\right)-2x+4034=0\)\(\Leftrightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)

    \(\Leftrightarrow2\left(x-2017\right)\cdot\left(4x-1\right)=0\)\(\Leftrightarrow\hept{\begin{cases}x-2017=0\\4x-1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)

          Vậy \(x=2017\)hoặc \(x=\frac{1}{4}\)

8x( x - 2017 ) - 2x + 4034 = 0

<=> 8x( x - 2017 ) - 2( x - 2017 ) = 0

<=> ( 8x - 2 )( x - 2017 ) = 0

<=> \(\orbr{\begin{cases}8x-2=0\\x-2017=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2017\end{cases}}\)

a) Thiếu VP

b) 4 - x = 2( x - 4 )²

<=> 4 - x = 2( x² - 8x + 16 )

<=> 4 - x = 2x² - 16x + 32

<=> 2x² - 16x + 32 - 4 + x = 0

<=> 2x² - 15x + 28 = 0

<=> 2x² - 8x - 7x + 28 = 0

<=> 2x( x - 4 ) - 7( x - 4 ) = 0

<=> ( x - 4 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}x-4=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{7}{2}\end{cases}}\)

c) ( x² + 1 )( x - 2 ) + 2x = 4

<=> x³ - 2x² + 3x - 2 - 4 = 0

<=> x³ - 2x² + 3x - 6 = 0

<=> x²( x - 2 ) + 3( x - 2 ) = 0

<=> ( x - 2 )( x² + 3 ) = 0

<=> x = 2 ( vì x² + 3 ≥ 3 > 0 ∀ x )

a, thiếu 

b, \(4-x=2\left(x-4\right)^2\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)

\(\Leftrightarrow4-x=2x^2-16x+32\Leftrightarrow2x^2-15x+28=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-7\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{7}{2}\end{cases}}\)

c, \(\left(x^2+1\right)\left(x-2\right)+2x=4\Leftrightarrow x^3-2x^2+3x-6=0\Leftrightarrow x_1=2;x_2=\sqrt{3}i\)

a) Ta có: \(8x\left(x-2017\right)-2x=4034\) (hẳn đề là ntn-.-)

\(\Leftrightarrow8x^2-16136x-2x-4034=0\)

\(\Leftrightarrow8x^2-16138x-4034=0\)

Bấm nghiệm ra: \(x_1=2017,499938\) ; \(x_2=-0,24993...\)

b) Ta có: \(4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+x-4=0\)

\(\Leftrightarrow2x^2-7x+4=0\)

\(\Leftrightarrow\left(x^2-\frac{7}{2}x+\frac{49}{16}\right)-\frac{17}{16}=0\)

\(\Leftrightarrow\left(x-\frac{7}{4}\right)^2-\left(\frac{\sqrt{17}}{4}\right)^2=0\)

\(\Leftrightarrow\left(x-\frac{7+\sqrt{17}}{4}\right)\left(x-\frac{7-\sqrt{17}}{4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7+\sqrt{17}}{4}\\x=\frac{7-\sqrt{17}}{4}\end{cases}}\)

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

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

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

Mà \(x^2+3\ge3>0\)

=> \(x-2=0\Rightarrow x=2\)

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

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

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

Bấm vào đúng 0 sẽ hiện ra kết quả đúng

g) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

  \(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

  \(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

  \(\Rightarrow-2\left(2x-5\right)=0\Rightarrow x=\dfrac{5}{2}\)

i) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

  \(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

  \(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)