Bài làm
\(\frac{50}{\left(20-x\right)}-\frac{50}{\left(x+20\right)}=\frac{4}{3}\)
\(\Leftrightarrow\frac{3.50\left(x+20\right)}{3.\left(20-x\right)\left(x+20\right)}-\frac{3.50\left(20-x\right)}{3\left(20-x\right)\left(x+20\right)}=\frac{4\left(20-x\right)\left(x+20\right)}{3\left(20-x\right)\left(x+20\right)}\)
\(\Rightarrow150\left(x+20\right)-150\left(20-x\right)=4\left(20^2-x^2\right)\)
\(\Leftrightarrow150x+3000-3000+150x=1600-4x^2\)
\(\Leftrightarrow4x^2+300x-1600=0\)
ĐKXĐ: \(x\notin\left\{-20;20\right\}\)
Ta có: \(\frac{50}{20-x}-\frac{50}{x+20}=\frac{4}{3}\)
\(\Leftrightarrow\frac{150\left(20+x\right)}{3\left(20-x\right)\left(20+x\right)}-\frac{150\left(20-x\right)}{3\left(20+x\right)\left(20-x\right)}-\frac{4\left(20+x\right)\left(20-x\right)}{3\left(20+x\right)\left(20-x\right)}=0\)
\(\Leftrightarrow300+150x-300+150x-4\left(400-x^2\right)=0\)
\(\Leftrightarrow300x-1600+4x^2=0\)
\(\Leftrightarrow4x^2-20x+320x-1600=0\)
\(\Leftrightarrow4x\left(x-5\right)+320\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4x+320\right)=0\)
\(\Leftrightarrow4\left(x-5\right)\left(x+80\right)=0\)
mà 4>0
nên \(\left[{}\begin{matrix}x-5=0\\x+80=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-80\left(tm\right)\end{matrix}\right.\)
Vậy: x\(\in\){-80;5}
