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a) (x-3)3-3+x=0
=> (x-3)3+(x-3)=0
=> (x-3)(x2-6x+10)
=> \(\left[{}\begin{matrix}x-3=0\\x^2-6x+10=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)
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\(\Rightarrow\left(x-4\right)\left(2x+x-4\right)=0\\ \Rightarrow\left(x-4\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)
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c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0
(\(x\) - 2022).(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)
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3(x+3)-x(x+3)=0
(x+3)(3-x) =0
x+3 =0 hoặc 3-x=0 =>x={-3;3}
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
y: Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
z: Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)
j: Ta có: \(25x^2-4=0\)
\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
x^2-x-2=0
x^2-2x+x-2=0
x(x-2)+(x-2)=0
(x+1)(x-2)=0
\(\Rightarrow\hept{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=2\end{cases}}}\)