Đáp án:
Giải thích các bước giải:
d)$\frac{1989.1990+3978}{1992.1991-3984}$
=$\frac{1989.1990+3978}{(1990+2).1991-3984}$
=$\frac{1989.1990+3978}{1990.1991+2.1991-3984}$
=$\frac{1989.1990+3978}{1990.1991+3982-3984}$
=$\frac{1989.1990+3978}{1990.1991-2}$
=$\frac{1989.1990+3978}{(1989+2).1990-2}$
=$\frac{1989.1990+3978}{1990.1989+2.1990-2}$
=$\frac{1989.1990+3978}{1990.1989+3980-2}$
=$\frac{1989.1990+3978}{1989.1990+3978}$
=1
a) $\frac{-315}{540}$ =$\frac{-7}{12}$
b) $\frac{25.13}{26.35}$ =$\frac{5.5.13}{13.2.7.5}$ =$\frac{5}{14}$
c) $\frac{6.9-2.17}{63.3-119}$ =$\frac{3.2.9-2.17}{9.7.3-7.17}$ =$\frac{2.(3.9-17)}{7.(3.9-17)}$ =$\frac{2}{7}$
