Ta có: 1.3.5.7...1991.1993 + 2.4.6......1992.1994 

= (1.3.5.7.19) . 9.11.....1991.1993 + (3.5.7.19).2.4.2.2.2.2.8....1992.1994

= 1995 . 9.11...1991.1993+1995.2.4.2.2.2.2.8....1992.1994

Vì cả 2 vế đều chia hết cho 1995 nên tổng chia hết cho 1995 

=(1.3.5.7.19).9.11..........1991.1993+(3.5.7.19).2.2.2.2.2.4.8......1992.1994

=1995.9.11....1991.1993+1995.2.2.2.2.2.4.8.....1992.1994

ca 2 so hang deu chia het cho1995 nen tong chia het cho 1995

\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}\\ =>\dfrac{x-5}{1990}-1+\dfrac{x-15}{1980}-1=\dfrac{x-1990}{5}-1+\dfrac{x-1980}{15}-1\\ =>\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}-\dfrac{x-1995}{5}-\dfrac{x-1995}{15}=0\\ =>\left(x-1995\right).\left(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{5}-\dfrac{1}{15}\right)=0\\ =>x-1995=0\\ =>x=1995\)

á đù đúng kwqo ưqffjoisdifadlsdfrgdfgdsfgr

Giải:

Ta gọi \(\dfrac{10^{1990}+1}{10^{1991}+1}\) =A và \(\dfrac{10^{1991}}{10^{1992}}\) =B

Ta có:

A=\(\dfrac{10^{1990}+1}{10^{1991}+1}\) 

10A=\(\dfrac{10^{1991}+10}{10^{1991}+1}\) 

10A=\(\dfrac{10^{1991}+1+9}{10^{1991}+1}\) 

10A=\(1+\dfrac{9}{10^{1991}+1}\) 

Tương tự:

B=\(\dfrac{10^{1991}}{10^{1992}}\) 

10B=\(\dfrac{10^{1992}}{10^{1992}}=1\) 

Vì \(\dfrac{9}{10^{1991}+1}< 1\) nên 10A<10B

⇒ \(\dfrac{10^{1990}+1}{10^{1991}+1}\) < \(\dfrac{10^{1991}}{10^{1992}}\)

đó bạn

đáng ra là toán lớp 6 đó nhưng mik thích đặt toán lớp 5 :)

A = \(\dfrac{10^{1990}+1}{10^{1991}+1}\) ⇒ 10A = \(\dfrac{10^{1991}+10}{10^{1991}+1}\) = \(1+\dfrac{9}{10^{1991}+1}\)

B =  \(\dfrac{10^{1991}+10}{10^{1992}+1}\) ⇒ 10B = \(\dfrac{10^{1992}+10}{10^{1992}+1}\) = 1 + \(\dfrac{9}{10^{1992}+1}\)

Vì \(\dfrac{9}{10^{1991}+1}\) > \(\dfrac{9}{10^{1992}+1}\)

10A > 10B => A > B