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

Câu 1: \(\frac{121.75.130.169}{39.60.11.198}\)=\(\frac{11x11x15x5x130x13x13}{13x13x15x4x11x11x18}\)= \(\frac{5x130}{4x18}\)= \(\frac{650}{72}\)= \(\frac{325}{36}\)

Câu 2: \(\frac{1998.1990+3978}{1992.1991-3984}\)= \(\frac{3976020}{3962088}\)= \(\frac{994005}{990522}\)

câu 2 bạn giải kĩ hơn đi.khó hiểu

Hình như bạn At the speed of light làm sai rồi . 39: 13 = 3 mà chứ đâu phải bằng 13. Bạn chưa học chia bao giờ à !!¤♡☆😮

\(a,\frac{121.75.130.169}{39.60.11.198}=\frac{11^2.5^2.3.13.2.5.13^2}{13.3.5.2^2.3.11.3^2.2.11}\)\(=\frac{11^2.5^3.3.13^3.2}{13.3^4.5.2^3.11^2}=\frac{5^2.13^2}{3^3.2^2}=\frac{4425}{108}\)

\(b,\frac{1989.1990+3978}{1992.1991-3984}=\frac{1989.1990+1989.2}{1992.1991-1992.2}\)\(=\frac{1989\left(1990+2\right)}{1992\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=\frac{1}{1}=1\)

\(c.\frac{135.350+135.550}{900.100+35.900}=\frac{135\left(350+550\right)}{900\left(100+35\right)}=\)\(\frac{135.900}{900.135}=\frac{1}{1}=1\)

\(d.\frac{243.650-243.350}{600.200+600.43}=\frac{243\left(650-350\right)}{600\left(200+43\right)}\)\(=\frac{243.300}{600.243}=\frac{300}{600}=\frac{1}{2}\)

$\frac{1989.1990+3978}{1992.1991-3984}$

= $\frac{1989.1990+198.2}{1992.1991-1992.2}$

= $\frac{1989.(1990=2)}{1992.(1991-2)}$

= $\frac{1989.1992}{1992.1989}$

= 1
 

Mik nghĩ phải là 1989 chứ...sao lại là 1998 nhỉ?

10 năm sau cũng chẳng có ai làm được

Rút gọn phân số:\(\frac{1989.1990+3978}{1992.1991-3984}\)

\(=\frac{1989.1990+1989.2}{1992.1991-1992.2}=\frac{1989.\left(1990+2\right)}{1992.\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=1\)

học giỏi nha

\(\frac{1989\cdot1990+3978}{1992\cdot1991-3984}\)

\(=\frac{1989\cdot1990+1989\cdot2}{1992\cdot1991-1992\cdot2}\)

\(=\frac{1989\cdot\left(1990+2\right)}{1992\cdot\left(1991-2\right)}\)

\(=\frac{1989\cdot1992}{1992\cdot1989}=1\)

cảm ơn