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a)
TH1: \(x< \dfrac{-2}{3}\)
<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.\)
PT <=> \(2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)\)
TH2: \(\dfrac{-2}{3}\le x< 4\)
<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)
PT <=> \(2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)\)
TH3: \(x\ge4\)
<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)
PT <=> \(0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)\)
KL: x \(\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}\)
b) TH1: \(x\ge-1< =>\left|x+1\right|=x+1\)
PT <=> 2x - x -1 = \(\dfrac{-1}{2}\)
<=> x = \(\dfrac{1}{2}\) (c)
TH2: x < -1 <=> \(\left|x+1\right|=-x-1\)
PT <=> 2x + x + 1 = \(\dfrac{-1}{2}\)
<=> x = \(\dfrac{-1}{2}\) (l)
KL: x \(\in\left\{\dfrac{1}{2}\right\}\)
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a) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Leftrightarrow\begin{cases}x+1< 0\\x-2>0\end{cases}\) hoặc \(\begin{cases}x+1>0\\x-2< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x< -1\\x>2\end{cases}\) (loại) hoặc \(\begin{cases}x>-1\\x< 2\end{cases}\)
\(\Leftrightarrow-1< x< 2\)
b)\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)
\(\Leftrightarrow\begin{cases}x-2>0\\x+\frac{2}{3}>0\end{cases}\) hoặc \(\begin{cases}x-2< 0\\x+\frac{2}{3}< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>2\\x>-\frac{2}{3}\end{cases}\) hoặc \(\begin{cases}x< 2\\x< -\frac{2}{3}\end{cases}\)
\(\Leftrightarrow x>2\) hoặc \(x< -\frac{2}{3}\)
a) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow x+1\) và \(x-2\) trái dấu nhau.
Mà \(x-2< x+1\) với mọi x
\(\Rightarrow\begin{cases}x-2< 0\\x+1>0\end{cases}\Leftrightarrow\begin{cases}x< 2\\x>-1\end{cases}\Leftrightarrow-1< x< 2\)
\(\Rightarrow x\in\left\{0;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-\frac{1}{2}\right)\left(y+\frac{1}{3}\right)\left(z-2\right)=0\) và \(x+2=y+3=z+4\)
\(\Rightarrow x-\frac{1}{2}=0\) hoặc \(y+\frac{1}{3}=0\) hoặc \(z-2=0\)
\(\Rightarrow x=\frac{1}{2}\) | \(y=-\frac{1}{3}\) | \(z=2\)
Khi \(x=\frac{1}{2}\) thì:
\(\frac{1}{2}+2=\frac{5}{2}\)
\(y=\frac{5}{2}-3=-\frac{1}{2}\)
\(z=\frac{5}{2}-4=\frac{-3}{2}\)
Khi \(y=\frac{-1}{3}\) thì:
\(\frac{-1}{3}+3=\frac{8}{3}\)
\(x=\frac{8}{3}-2=\frac{2}{3}\)
\(z=\frac{8}{3}-4=-\frac{4}{3}\)
Khi \(z=2\) thì:
\(2+4=6\)
\(x=6-2=4\)
\(y=6-3=3\)
Vậy (x,y,z) = \(\left(\frac{1}{2};-\frac{1}{2};-\frac{3}{2}\right)\) ; \(\left(\frac{2}{3};-\frac{1}{3};-\frac{4}{3}\right)\) ; \(\left(4;3;2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Để M<0 thì:
\(\hept{\begin{cases}x-1< 0\\x+2< 0\\3-x< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 1\\x-2\\x>3\end{cases}}\)
Không chắc lắm đâu
#Châu's ngốc
![](https://rs.olm.vn/images/avt/0.png?1311)
Lớp 7 cần lập bảng ra; các điểm quan trọng
x={-2,1,3
cách khác,
\(\Leftrightarrow-M=\left(x-1\right)\left(x+2\right)\left(x-3\right)>0\)
\(x< -2\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x+2< 0\\x-3< 0\\-M< 0\Rightarrow M>0\Rightarrow.vN_o\end{matrix}\right.\)
\(-2< x< 1\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x+2>0\\x-3< 0\\-M>0\Rightarrow M< 0\Rightarrow.N_o:-2< x< 1\end{matrix}\right.\)
\(1< x< 3\Rightarrow\left\{{}\begin{matrix}x-1>0\\x+2>0\\x-3< 0\\-M< 0\Rightarrow M>0\Rightarrow.vN_o\end{matrix}\right.\)
\(x>3\Rightarrow\left\{{}\begin{matrix}x-1>0\\x+2>0\\x-3>0\\-M>0\Rightarrow M< 0\Rightarrow.vN_o:x>3\end{matrix}\right.\)
Kết luận: \(\left[{}\begin{matrix}1< x< 2\\x>3\end{matrix}\right.\)
đổi -M để cho các nhân tử(x-1)(x+2)(x-3) cùng chiều x đỡ nhầm