$3x^2-7x+4=0$
⇔ $3x^2-3x-4x+4=0$
⇔ $3x(x-1)-4(x-1)=0$
⇔ $(x-1)(3x-4)=0$
⇔ \(\left[ \begin{array}{l}x-1=0\\3x-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=\frac{4}{3}\end{array} \right.\)
Vậy S = {$1;\frac{4}{3}$}
`\frac{x+4}{5-x}-5=\frac{x+3}{10}-\frac{x-1}{3}` (có sai đề không bạn?)
⇒ $30(x+4)-150(5-x)=3(x+3)(5-x)-10(x-1)(5-x)$
⇔ $30x+120-750+150x=3(5x-x^2+15-3x)-10(5x+x^2-5+x)$
⇔ $30x+120-750+150x=15x-3x^2+45-9x-50x-10x^2+50-10x$
⇔ $30x+150x-15x+3x^2+9x+50x+10x^2+10x=45+50-120+750$
⇔ $224x+13x^2=725$
⇔ Bấm nghiệm bằng máy tính.
