tìm xEQ, biết:
a)(x+1)(x-2)<0
b)(x-2)(x+2/3)>0
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a, |x^2 - 3x| = 0
=> x^2 - 3x = 0
=> x(x - 3) = 0
=> x = 0 hoặc x - 3 = 0
=> x = 0 hoặc x = 3
vậy_
\(\left|a^2-3a\right|=0\)
\(\Rightarrow a^2-3a=0\)
\(\Rightarrow a\left(a-3\right)=0\)
\(\Rightarrow\hept{\begin{cases}a=0\\a=3\end{cases}}\)
a) (-3).(x+2)<0
=>x+2>0
=>x> -2
b)(x-1).(x+\(\dfrac{1}{3}\))>0
<=>(x-1) và \(\left(x+\dfrac{1}{3}\right)\) cùng dấu
TH1: x-1 <0
và x+\(\dfrac{1}{3}\)<0
\(\left\{{}\begin{matrix}x< 1\\x< \dfrac{-1}{3}\end{matrix}\right.\) =>x<\(\dfrac{-1}{3}\)
TH2:x-1>0
và x+\(\dfrac{1}{3}\)>0
\(\left\{{}\begin{matrix}x>1\\x>\dfrac{-1}{3}\end{matrix}\right.\)=>x>1
a)(-3).(x+2) <0
=> x+2> 0
=> x>2
b)(x-1).(x+\(\dfrac{1}{3}\)) >0
=> x-1>0 hay x+\(\dfrac{1}{3}\) >0
=> x>1hay x>-\(\dfrac{1}{3}\)
a) \(\sqrt{x}\left(\sqrt{x}-1\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
b) \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\\sqrt{x}=-3\left(vôlí\right)\end{cases}}\)
c) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=0\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\\sqrt{x}+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(vôlí\right)\\\sqrt{x}=-3\left(vôlí\right)\end{cases}}\)
Bài 2:
a, |x-1| -x +1=0
|x-1| = 0-1+x
|x-1| = -1 + x
\(\orbr{\begin{cases}x-1=-1+x\\x-1=1-x\end{cases}}\)
\(\orbr{\begin{cases}x=-1+x+1\\x=1-x+1\end{cases}}\)
\(\orbr{\begin{cases}x=x\\x=2-x\end{cases}}\)
x = 2-x
2x = 2
x = 2:2
x=1
b, |2-x| -2 = x
|2-x| = x+2
\(\orbr{\begin{cases}2-x=x+2\\2-x=2-x\end{cases}}\)
2-x = x+2
x+x = 2-2
2x = 0
x = 0
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a) \(\left(x+1\right)\left(x-2\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-1< x< 2\\x\in\varnothing\end{matrix}\right.\) vậy \(-1< x< 2\)
b) \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>2\) hoặc \(x< \dfrac{-2}{3}\)