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a) Ta có: \(\left|x-3\right|+\left|y-2x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\y-2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x=2\cdot3=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2\(x\) - 1).(2\(x\) - 5) < 0
Lập bảng ta có:
\(x\) | \(\dfrac{1}{2}\) \(\dfrac{5}{2}\) |
2\(x\) - 1 | - 0 + + |
2\(x\) - 5 | - - 0 + |
(2\(x\) - 1).(2\(x\) - 5) | + 0 - 0 + |
Theo bảng trên ta có: \(\dfrac{1}{2}\) < \(x\) < \(\dfrac{5}{2}\)
(3 - 2\(x\)).(\(x\) + 2) > 0
Lập bảng ta có:
\(x\) | -2 \(\dfrac{3}{2}\) |
3 - 2\(x\) | + + 0 - |
\(x\) + 2 | - 0 + + |
(3 -2\(x\)).(\(x\) +2) | - 0 + 0 - |
Theo bảng trên ta có: - 2 < \(x\) < \(\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : (x + 1)(x - 3) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Ta có : \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\-\frac{1}{2}x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=1\\-\frac{1}{2}x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-5.\left(-2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(3x+5)(2x-7)=0
\(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-5\\2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-5}{3}\\x=\frac{7}{2}\end{matrix}\right.\)
(-5x+2)(-3x-4)=0
\(\Leftrightarrow\left[{}\begin{matrix}\left(-5x+2\right)=0\\\left(-3x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=-2\\-3x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{5}\\x=\frac{-3}{4}\end{matrix}\right.\)
(x-5)(4x-3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{4}\end{matrix}\right.\)
-2x(x+1)(x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)
\(\left(3x+5\right).\left(2x-7\right)=0\)
=> \(\left\{{}\begin{matrix}3x+5=0\\2x-7=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}3x=0-5=-5\\2x=0+7=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\left(-5\right):3\\x=7:2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-\frac{5}{3}\\x=\frac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{2}\right\}\).
\(\left(-5x+2\right).\left(-3x-4\right)=0\)
=> \(\left\{{}\begin{matrix}-5x+2=0\\-3x-4=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}-5x=0-2=-2\\-3x=0+4=4\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=\left(-2\right):\left(-5\right)\\x=4:\left(-3\right)\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\frac{2}{5}\\x=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{2}{5};-\frac{4}{3}\right\}\).
Mấy câu còn lại bạn làm tương tự nhé.
Chúc bạn học tốt!
TH1 : x + 5 = 0
x = 0 - 5
x = -5
TH2 : 1 - 2x = 0
2x = 1-0
2x = 1
x = 1 : 2
x = 1/2
Vậy x thuộc { -5 ; 1/2 }
Ta có : | x+5| \(\ge\) 0 ; |1-2x| \(\ge\) 0
=> |x+5 | + |1-2x | \(\ge\) 0
mà |x+5 | + |1-2x | = 0
=> \(\hept{\begin{cases}x+5=0\\1-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=\frac{1}{2}\end{cases}}}\)
Vây x =-5 ; x=1/2