a) * Nếu 4x - 5 \(\ge\) 0 thì x \(\ge\) \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(3-2\left(4x-5\right)=\dfrac{2}{6}\)
\(\Leftrightarrow\) \(-8x=-3-10+\dfrac{2}{6}\)
\(\Leftrightarrow\) x = \(\dfrac{19}{12}\) (t/m)
* Nếu 4x - 5 < 0 thì x < \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(3-2\left(-4x+5\right)=\dfrac{2}{6}\)
\(\Leftrightarrow\) \(3+8x-10=\dfrac{2}{6}\)
\(\Leftrightarrow\) x = \(\dfrac{11}{12}\) (t/m)
b) Không hiểu đề :v
c) \(\left(7-3x\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d) \(2x\left(5-3x\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow0< x< \dfrac{5}{3}\)
e) \(\left(4-2x\right)\left(5x+3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-2x< 0\\5x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4-2x>0\\5x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< -\dfrac{3}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x>-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
Loại TH1, nhận TH2
Vậy \(-\dfrac{3}{5}< x< 2\)
g) \(\left|3x+1\right|+\left|1-3x\right|=0\) (1)
* Nếu x < \(\dfrac{-1}{3}\)
PT (1) \(\Leftrightarrow-3x-1-1+3x=0\)
0x - 2 = 0
0x = 2 \(\Rightarrow\) PT vô nghiệm
* Nếu \(\dfrac{-1}{3}\le x\le\dfrac{1}{3}\)
PT (1) \(\Leftrightarrow3x+1-1+3x=0\)
6x = 0
x = 0 (t/m)
* Nếu x > \(\dfrac{1}{3}\)
PT (1) \(\Leftrightarrow3x+1+1-3x=0\)
0x + 2 = 0
0x = -2
PT vô nghiệm.
Vậy x = 0
a, \(3-2\left|4x-5\right|=\dfrac{2}{6}\)
\(\Rightarrow2\left|4x-5\right|=\dfrac{8}{3}\)
\(\Rightarrow\left|4x-5\right|=\dfrac{4}{3}\)
+) Xét \(x\ge\dfrac{5}{4}\) có:
\(4x-5=\dfrac{4}{3}\Rightarrow4x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{12}\) ( t/m )
+) Xét \(x< \dfrac{5}{4}\) có:
\(4x-5=\dfrac{-4}{3}\Rightarrow4x=\dfrac{11}{3}\Rightarrow x=\dfrac{11}{12}\) ( t/m )
Vậy...
b, tương tự
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
d, \(2x\left(5-3x\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x< 0\\5-3x< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{3}{5}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x>\dfrac{3}{5}\end{matrix}\right.\) (loại )
Vậy \(0< x< \dfrac{3}{5}\)
e, tương tự
g, \(\left|3x+1\right|+\left|1-3x\right|=0\)
\(\Rightarrow\left|3x+1\right|+\left|3x-1\right|=0\)
+) Xét \(x\ge\dfrac{1}{3}\) có:
\(3x+1+3x-1=0\)
\(\Rightarrow6x=0\)
\(\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-1}{3}\le x< \dfrac{1}{3}\) có:
\(3x+1+1-3x=0\)
\(\Rightarrow2=0\) ( vô lí )
+) Xét \(x< \dfrac{-1}{3}\) có:
\(-3x-1+1-3x=0\)
\(\Rightarrow-6x=0\Rightarrow x=0\) ( ko t/m )
Vậy ko có giá trị x thỏa mãn đề bài
