Sửa đề: \(C=3\dfrac{1}{117}.4\dfrac{1}{119}-1\dfrac{116}{117}.5\dfrac{118}{119}+\dfrac{5}{119}-\dfrac{10}{117}\)
\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(1+1-\dfrac{1}{117}\right)\left(5+1-\dfrac{1}{110}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)
\(=\left(3+\dfrac{1}{117}\right)\left(4+\dfrac{1}{119}\right)-\left(2-\dfrac{1}{117}\right)\left(6-\dfrac{1}{119}\right)+5.\dfrac{1}{119}-10.\dfrac{1}{117}\)
Đặt \(a=\dfrac{1}{117}\) và \(b=\dfrac{1}{119}\) ta có:
\(C=\left(3+a\right).\left(4+b\right)-\left(2-a\right)\left(6-b\right)+5b-10a\)
\(=12+3b+4a+ab-12+2b+6a-ab+5b-10a\)
\(=10b=10.\dfrac{1}{119}=\dfrac{10}{119}\)
