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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
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
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
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: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
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
a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
=>\(13\sqrt{2x}=28\)
=>căn 2x=28/13
=>2x=784/169
=>x=392/169
b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>2*căn x-5=4
=>căn x-5=2
=>x-5=4
=>x=9
c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=>x-2=0 hoặc x+2=1
=>x=-1 hoặc x=2
2 8