Ai làm đúng và nhanh nhất mk k cho

cái thứ hai lớn hơn

Ta có : \(\frac{1212+1313+1414+1515+1616}{1717+1818+1919+2020+2121}=\frac{12+13+14+15+16}{17+18+19+20+21}=\frac{28\cdot\frac{5}{2}}{38\cdot\frac{5}{2}}=\frac{14}{19}\)

Dễ thấy \(\frac{111111}{666665}< \frac{1}{2}\) ( do 111111 < 666665/2 ) và \(\frac{14}{19}>\frac{1}{2}\) ( do 14 > 19/2 )

Vậy \(\frac{111111}{666665}< \frac{1212+1313+1414+1515+1616}{1717+1818+1919+2020+2121}\)

1313/1515=1313:101/1515:101=13/15

vay 13/15=1313/1515

Ta có:1313/1515=13.101/15.101=13/15

Vì 13/15=13/15 nên 13/15=1313/1515

\(\dfrac{1111}{1212}\)  và  \(\dfrac{1515}{1616}\)

\(\dfrac{1111}{1212}=\dfrac{1111\times1616}{1212\times1616}=\dfrac{1795376}{1958592}\)

\(\dfrac{1515}{1616}=\dfrac{1515\times1212}{1616\times1212}=\dfrac{1836180}{1958592}\)

Vì: 1836180 > 1795376

=> \(\dfrac{1111}{1212}< \dfrac{1515}{1616}\)

\(\dfrac{1111}{1212}\) = \(\dfrac{1111:101}{1212:101}\) = \(\dfrac{11}{12}\)= 1 - \(\dfrac{1}{12}\) ;    \(\dfrac{1515}{1616}\) =\(\dfrac{1515:101}{1616:101}\) = \(\dfrac{15}{16}\) = 1 -\(\dfrac{1}{16}\)

Vì \(\dfrac{1}{12}\) > \(\dfrac{1}{16}\) nên  \(\dfrac{1111}{1212}\) < \(\dfrac{1515}{1616}\)

\(\dfrac{1111}{1212}=\dfrac{11}{12}=1-\dfrac{1}{12}\)

\(\dfrac{1516}{1516}=\dfrac{15}{16}=1-\dfrac{1}{16}\)

Vì: \(\dfrac{1}{12}>\dfrac{1}{16}\) nên \(1-\dfrac{1}{12}< 1-\dfrac{1}{16}\)

Vậy\(\dfrac{1111}{1212}< \dfrac{1515}{1616}\)

Trước khi so sánh, ta phải rút gọn phân số \(\frac{1212}{1515}\)và \(\frac{12}{15}\)

Rút gọn P/S, ta có: \(\frac{1212:101}{1515:101}\)=\(\frac{12:3}{15:3}\)=\(\frac{3}{5}\)(1)

                                   \(\frac{12:3}{15:3}=\frac{3}{5}\)(2)

Từ (1) và (2) => \(\frac{12}{15}=\frac{1212}{1515}\)

Ta có: \(\frac{12}{15}=\frac{12\div3}{15\div3}=\frac{4}{5}\)

\(\frac{1212}{1515}=\frac{1212\div303}{1515\div303}=\frac{4}{5}\)

Mà \(\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow\frac{12}{15}=\frac{1212}{1515}\)

Ta có :

 1212/1515=12.102/15.101=12/15

 Vì 12/15=12/15 =>12/15=1212/1515

vậy ...........

    k mk nha

DẤu nhân không phải x đâu nhe!

Cả hai đều sai 

Câu1 : Đ

Câu 2 : S

Ta có: \(\frac{1999x2000}{1999x2000+1}=\frac{1999x2000+1-1}{1999x2000+1}=1-\frac{1}{1999x2000+1}\)

\(\frac{2000x2001}{2000x2001+1}=\frac{2000x2001+1-1}{2000x2001+1}=1-\frac{1}{2000x2001+1}\)

Nhận thấy: \(\frac{1}{1999x2000+1}>\frac{1}{2000x2001+1}\)=> \(1-\frac{1}{1999x2000+1}< 1-\frac{1}{2000x2001+1}\)

=> \(\frac{1999x2000}{1999x2000+1}=\frac{2000x2001}{2000x2001+1}\)

\(\frac{1999x2000}{1999x2000+1}< \frac{2000x2001}{2000x2001+1}\)