Question

a, (3x - 2/4) . (x + 1/2) = 0

b, (2x - 5) . ( 3/2x +9) .(0,3x - 12) = 0

Answer 1

a/ \(\left(3x-\dfrac{2}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy ................

b/ \(\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)

Vậy ..

Answer 2

\(a)\left(3x-\dfrac{2}{4}\right).\left(x+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

\(b)\left(2x-5\right).\left(\dfrac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\dfrac{3}{2}x+9=0\\0,3x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\\dfrac{3}{2}x=-9\\0,3x=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-6\\x=40\end{matrix}\right.\)

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