Question

a) (2x-1)(3-2x)=0

b) x(x+1)(x+\(\dfrac{3}{4}\))=0

c) (\(\dfrac{1}{2}\)-x)(\(\dfrac{3}{4}\)-\(\dfrac{1}{2}\)x)=0

d) (\(X^2\)-1)(2x-1)=(\(X^2\)-1)(x+3)

Answer 1

Answer 2

a: =>2x-1=0 hoặc 3-2x=0

=>x=1/2 hoặc x=3/2

b: \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+\dfrac{3}{4}=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1;-\dfrac{3}{4}\right\}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=0\\\dfrac{3}{4}-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{2}x=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{1}{2};\dfrac{3}{2}\right\}\)

d: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1-x-3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-4\right)=0\)

hay \(x\in\left\{1;-1;4\right\}\)