Question

Giải phương trình sau :

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

Answer 1

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

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

\(3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\) ;\(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}=0\Leftrightarrow\frac{2\left(x+3\right)}{7}=\frac{4x-3}{5}\Leftrightarrow10\left(x+3\right)=7\left(4x-3\right)\Leftrightarrow x=\frac{17}{6}\).

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{2}{3};\frac{7}{16}\right\}\).

Answer 2

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\10\left(x+3\right)=7\left(4x-3\right)\end{cases}}\)

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

vậy x=2/3 hoặc x=17/6

Answer 3

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

\(\Leftrightarrow\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\frac{5\left(2x+6\right)}{7.5}-\frac{7\left(4x-3\right)}{5.7}\right)=0\)

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

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

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

Vậy....

Answer 4

Đính chính : Ở chỗ tập nghiệm sai có ghi là : \(S=\left\{\frac{2}{3};\frac{7}{16}\right\}\) xin sửa lại là :\(S=\left\{\frac{2}{3};\frac{17}{6}\right\}\).

Ok !