a, \(x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow x=0;x=-1\)
b, \(x\left(x-3\right)+4x-12=0\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\Leftrightarrow x=3;x=-4\)
c, \(2x\left(x-2\right)+3\left(x-2\right)=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{3}{2};x=2\)
d, \(5x\left(x-2023\right)-x+2023=0\Leftrightarrow5x\left(x-2023\right)-\left(x-2023\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2023\right)=0\Leftrightarrow x=\dfrac{1}{5};x=2023\)
a: \(x^2+x=0\)
=>x(x+1)=0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
b: \(x\left(x-3\right)+4x-12=0\)
=>x(x-3)+4(x-3)=0
=>(x-3)(x+4)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
c: \(2x\left(x-2\right)+3\left(x-2\right)=0\)
=>(x-2)(2x+3)=0
=>\(\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: 5x(x-2023)-x+2023=0
=>5x(x-2023)-(x-2023)=0
=>(5x-1)(x-2023)=0
=>\(\left[{}\begin{matrix}5x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2023\end{matrix}\right.\)
`a)x^2+x=0`
`<=>x(x+1)=0`
`<=>[(x=0),(x=-1):}`
`b)x(x-3)+4x-12=0`
`<=>x(x-3)+4(x-3)=0`
`<=>(x-3)(x+4)=0`
`<=>[(x=3),(x=-4):}`
`c)2x(x-2)+3(x-2)=0`
`<=>(x-2)(2x+3)=0`
`<=>[(x=2),(x=-3/2):}`
`d)5x(x-2023)-x+2023=0`
`<=>(x-2023)(5x-1)=0`
`<=>[(x=2023),(x=1/5):}`
`#3107.101107`
`4.`
`a)`
`x^2 + x = 0`
`=> x(x + 1) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy, `x \in {0; -1}`
`b)`
\(x\left(x-3\right)+4x-12=0\)
`=> x(x - 3) + 4(x - 3) = 0`
`=> (x + 4)(x - 3) = 0`
`=>`\(\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-4\\x=3\end{matrix}\right.\)
Vậy, `x \in {-4; 3}`
`c)`
\(2x\left(x-2\right)+3\left(x-2\right)=0\)
`=> (2x + 3)(x - 2) = 0`
`=>`\(\left[{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\x=2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=2\end{matrix}\right.\)
Vậy, `x \in {-3/2; 2}`
`d)`
\(5x\left(x-2023\right)-x+2023=0\)
`=> 5x(x - 2023) - (x - 2023) = 0`
`=> (5x - 1)(x - 2023) = 0`
`=>`\(\left[{}\begin{matrix}5x-1=0\\x-2023=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}5x=1\\x=2023\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2023\end{matrix}\right.\)
Vậy, `x \in {1/5; 2023}.`
