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

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

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

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

 

18 tháng 7 2023

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

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

 

 

28 tháng 8 2020

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

<=> x- 2x2 + 3x - 6 = 0

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

<=> ( x - 2 )( x2 + 3 ) = 0

<=> x = 2 ( vì x2 + 3 ≥ 3 > 0 ∀ x )

28 tháng 8 2020

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: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)

\(\Leftrightarrow-2x+1-x-2=8\)

\(\Leftrightarrow-3x=9\)

hay x=-3

b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)

\(\Leftrightarrow-2x=-2\)

hay x=1

17 tháng 10 2020
https://i.imgur.com/KdSDH1v.jpg
29 tháng 8 2021

\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

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

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

29 tháng 8 2021

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

vậy ...

 

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

    \(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

31 tháng 8 2021

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

31 tháng 8 2021

Sorry, cái này mình nhầm