Question

Giải phương trình :

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

Answer 1

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

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\end{cases}}\)

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

\(\Leftrightarrow5.2\left(x+3\right)=7\left(4x-3\right)\)

\(\Leftrightarrow10x+30=28x-21\)

\(\Leftrightarrow10x-28x=-21-30\)

\(\Leftrightarrow-18x=-51\)

\(\Leftrightarrow x=\frac{17}{6}\)

Answer 2

sex em ko

Answer 3

Giải :

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

\(\Leftrightarrow3x-2=0\text{ hoặc }\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\)

TH1 : \(3x-2=0\)

          \(\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\).

TH2 : \(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\)

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

          \(\Leftrightarrow10\left(x+3\right)=7\left(4x-3\right)\)

          \(\Leftrightarrow x=\frac{17}{6}\).

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