Ta có:
\(\frac{121.75.130.169}{39.60.11.198}=\frac{11^2.3.5^2.2.5.13.13^2}{3.13.2^2.3.5.11.2.3^2.11}=\frac{11^2.3.5^3.2.13^3}{3^4.13.2^3.5.11^2}=\frac{5^2.13^2}{3^3.2^2}=\frac{4225}{108}\)
\(\frac{1998.1990+3978}{1992.1991-3984}=\frac{2.3^3.37.2.5.199+2.3^2.13.17}{2^3.3.83.11.181-2^4.3.83}=\frac{2.3^2\left(3.37.2.5.199+13.17\right)}{2^3.3.83\left(11.181-2\right)}=\frac{3.221111}{2^2.83.1989}\)
b)\(\frac{1989.1990+3978}{1992.1991-3984}=\frac{1989.1990+1989.2}{1992.1991-1992.2}=\frac{1989.\left(1990+2\right)}{1992.\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=1\)
Chúc bn học tốt
a)\(\frac{121.75.130.169}{39.60.11.198}=\frac{11^2.3.5^2.2.5.13.13^2}{3.13.2^2.3.5.11.2.3^2.11}=\frac{2.3.5^3.11^2.13^3}{2^3.3^4.5.11^2.13}\)\(=\frac{5^2.13^2}{2^2.3^3}=\frac{25.169}{4.27}=\frac{4225}{108}=39\frac{13}{108}\)
b)\(\frac{1998.1990+3978}{1992.1991-3984}\)đề có sai không bạn bn nói nhanh mk làm phần b cho
Chúc bn học tốt
