a: =>x-1/2=0 hoặc 2x+5=0
=>x=1/2 hoặc x=-5/2
b:=>-4x=-6
=>x=3/2
\(\left(x-\dfrac{1}{2}\right).\left(2x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}\\2x+5=0\end{matrix}\right.=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{-5}{2}\right\}\)
\(b,15-7x=9-3x\\ \Leftrightarrow-7x+3x=-15+9\\ \Leftrightarrow-4x=-6\\ \Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(x=\dfrac{3}{2}\)
a)\(\left(\dfrac{x-1}{2}\right)\left(2x-5\right)=0\)
⇔\(\dfrac{x-1}{2}=0\) \(hoặc\) \(2x-5=0\)
⇔\(\dfrac{x-1}{2}\cdot\dfrac{1}{2}=0\cdot\dfrac{1}{2}\) \(hoặc\) \(2x=5\)
⇔\(x-1=0\) \(hoặc\) \(\dfrac{2x}{2}=\dfrac{5}{2}\)
⇔\(x=1\) \(hoặc\) \(x=\dfrac{5}{2}\)
Vậy phương trình có tập nghiệm là: \(S=\left\{1;\dfrac{5}{2}\right\}\)
b)\(15-7x=9-3x\)
⇔\(-7x+3x=9-15\)
⇔\(-4x=-6\)
⇔\(\dfrac{-4x}{-4}=\dfrac{-6}{-4}\)
⇔\(x=\dfrac{-6}{-4}\)
⇔\(x=\dfrac{3}{2}\)
Vậy phương trình có nghiệm là \(x=\dfrac{3}{2}\)
