Question

[Uchiha Sasuke]

Giải pt sau :

\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)

Answer 1

[๖Fly༉Donutღღ]

\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)

\(\left(3x-2\right).\frac{10\left(x+3\right)-7\left(4x-3\right)}{35}=0\)

\(\left(3x-2\right)\left(10x+30-28x+21\right)=0\)

\(\left(3x-2\right)\left(51-18x\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-2=0\\51-18x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\-18x=-51\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}}\)

Vậy \(S=\left\{\frac{2}{3};\frac{17}{6}\right\}\)

Answer 2

[Dương Lam Hàng]

\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)

\(\Leftrightarrow\left(3x-2\right)\left[\frac{2.5\left(x+3\right)}{35}-\frac{7\left(4x-3\right)}{35}\right]=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\\frac{-18x+51}{35}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\-18x+51=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}}\)

Vậy \(x=\left\{\frac{2}{3};\frac{17}{6}\right\}\)