Gọi x là số khoa ban đầu ở gánh hàng của Nam (đk: x nguyên dương)
Theo đề ta có pt:
\(x+\left(\dfrac{1}{4}+3\right)-\left[\dfrac{x-\left(\dfrac{1}{4}+3\right)}{3}+4\right]-\left[\dfrac{x-\left(\dfrac{1}{4}+3\right)-\left\{\dfrac{x-\dfrac{1}{4}+3}{3}+4\right\}}{2}+5\right]=5\)
\(\Leftrightarrow x+\dfrac{13}{4}-\left[\dfrac{x-\dfrac{13}{4}}{3}+4\right]-\left[\dfrac{x-\dfrac{13}{4}-\left(\dfrac{x-\dfrac{13}{4}}{3}+4\right)}{2}+5\right]=5\)
\(\Leftrightarrow x+\dfrac{13}{4}-\left(\dfrac{x-\dfrac{13}{4}}{3}+\dfrac{12}{3}\right)-\left[\dfrac{x-\dfrac{13}{4}-\left(\dfrac{x-\dfrac{13}{4}}{3}+\dfrac{12}{3}\right)}{2}+5\right]=5\)
\(\Leftrightarrow x+\dfrac{13}{4}-\left(\dfrac{x-\dfrac{13}{4}+12}{3}\right)-\left[\dfrac{x-\dfrac{13}{4}-\left(\dfrac{x-\dfrac{13}{4}+12}{3}\right)}{2}+\dfrac{10}{2}\right]=5\)
\(\Leftrightarrow x+\dfrac{13}{4}-\left(\dfrac{x+\dfrac{35}{4}}{3}\right)-\left[\dfrac{x-\dfrac{13}{4}-\left(\dfrac{x+\dfrac{35}{4}}{3}\right)}{2}+\dfrac{10}{2}\right]=5\)
\(\Leftrightarrow x+\dfrac{13}{4}-\dfrac{x+\dfrac{35}{4}}{3}-\left[\dfrac{x-\dfrac{13}{4}-\dfrac{x+\dfrac{35}{4}}{3}}{2}+\dfrac{10}{2}\right]=5\)
\(\Leftrightarrow x+\dfrac{13}{4}-\dfrac{x+\dfrac{35}{4}}{3}-\left(\dfrac{x-\dfrac{13}{4}-\dfrac{x+\dfrac{35}{4}}{3}+10}{2}\right)=5\)
\(\Leftrightarrow\dfrac{12x}{12}+\dfrac{4\left(x+\dfrac{35}{4}\right)}{12}-\dfrac{6\left(x-\dfrac{13}{4}-\dfrac{x+\dfrac{35}{4}}{3}+10\right)}{12}=5\)
\(\Leftrightarrow\dfrac{12x}{12}+\dfrac{4x+35}{12}-\dfrac{6x-19,5-\dfrac{6x+52,5}{3}+60}{12}=5\)
\(\Leftrightarrow\dfrac{12x}{12}+\dfrac{4x+35}{12}-\dfrac{\dfrac{18x}{3}-\dfrac{58,5}{3}-\dfrac{6x+52,5}{3}+\dfrac{180}{3}}{12}=5\)
\(\Leftrightarrow\dfrac{12x}{12}+\dfrac{4x+35}{12}-\dfrac{\dfrac{18x-58,5-6x-52,5+180}{3}}{12}=5\)
\(\Leftrightarrow\dfrac{12x}{12}+\dfrac{4x+35}{12}-\dfrac{12x+69}{12}-5=0\)
\(\Leftrightarrow\dfrac{12x}{12}+\dfrac{4x+35}{12}-\dfrac{12x+69}{12}-\dfrac{60}{12}=0\)
\(\Leftrightarrow12x+4x-35-12x-69-60=0\)
\(\Leftrightarrow4x-164=0\)
\(\Leftrightarrow4x=164\)
\(\Rightarrow x=\dfrac{164}{4}=41\)
Vậy....
(Cứ từ từ mà ngẫm:"))
