Question

\(\left(\dfrac{-1}{2}+3x\right)\left(1-\dfrac{2}{3}x\right)1999=0\)
giải pt

Answer 1

\(A=\left(\dfrac{-1}{2}+3x\right)\left(1-\dfrac{2}{3}x\right)\cdot1999=0\)

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

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

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{3}{2}\end{matrix}\right.\)thì GTBT trên bằng 0.

Answer 2

\(\left(\dfrac{-1}{2}+3x\right)\left(1-\dfrac{2}{3}x\right)1999=0\)\(\Leftrightarrow\left(\dfrac{-1}{2}+3x\right)\left(1-\dfrac{2}{3}x\right)=0\)

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

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

Vậy phương trình có nghiệm \(x=\dfrac{3}{2}\)

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