Lời giải:

$\frac{-5071}{3333}=-1,52145< -1,5213$

Gọi phân số cần tìm có dạng $\frac{11}{a}$. Đương nhiên, $a< 0$ và $a$ nguyên.

Ta có:

$\frac{-13}{2}< \frac{11}{a}< \frac{-13}{3}$

\(\Rightarrow 22< -13a< 33\)

\(\Leftrightarrow \frac{-22}{13}> a> \frac{-33}{13}\)

Vì $a$ nguyên nên $a=-2$

Do đó phân số cần tìm là: $\frac{11}{-2}$

ta có :  \(1+\frac{-33}{19}=\frac{-14}{19}\)

\(1+\frac{-45}{31}=\frac{-14}{31}\)

Vì 19 < 31 Nên \(\frac{-14}{19}>\frac{-14}{31}\)

Vậy : \(\frac{-33}{19}< \frac{-45}{31}\)

Bài 1 : 

a) \(-\frac{33}{19}\) và \(\frac{-45}{31}\)

ta có : \(-\frac{31}{19}\) +1=\(\frac{-14}{19}\)

             \(\frac{-41}{31}\)+1=\(\frac{-14}{31}\)

vì 19<31 =>\(\frac{-14}{19}\) > \(\frac{-14}{31}\)

Vậy \(\frac{-31}{19}\) > \(\frac{-41}{31}\)

Bài 1 :

b)-1,5213 và \(-\frac{5017}{3333}\)

Ta có:  -1,5213=\(\frac{-15213}{10000}\)

              \(\frac{-5017}{3333}\)=\(\frac{-15051}{9999}\) >\(\frac{-15051}{10000}\)

vì -15213<-15051 và 10000>0

=>\(\frac{-15213}{10000}\) < \(\frac{-15051}{10000}\) < \(\frac{-15051}{9999}\)

Vậy -1,5213 < \(\frac{-5071}{3333}\)

a,4/9>3/10

b, 11/19<13/18

c, 45/97<47/96

d,20/31>19/33

e,45/46<44/47

bạn ơi có thể g cho mình ko quy dong hay so sánh phan so 9/10 va 10/11

\(M=\frac{19^{30}+5}{19^{31}+5}\)

\(19M=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5}{19^{31}+5}+\frac{90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(N=\frac{19^{31}+5}{19^{32}+5}\)

\(19N=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5}{19^{32}+5}+\frac{90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

chung tử rồi so sánh mẫu đi

#)Giải :

\(M=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19M=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)

\(N=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19N=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)

Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow19M>19N\Rightarrow M>N\)

              #~Will~be~Pens~#

Ta có : \(N=\frac{19^{31}+5}{19^{32}+5}< 1\)

Áp dụng công thức \(\forall a,b,m\in N;b,m\inℕ^∗\)

\(\Rightarrow\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

Ta có :

\(N=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\cdot\left(19^{30}+5\right)}{19\cdot\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=M\)

Vậy N < M

ớ chết, mk nhầm, lm lại nha

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)

\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)

=> \(S< \frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(S< 30.\frac{1}{60}\)

\(S< \frac{1}{2}< \frac{4}{5}\)

\(S< \frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Rightarrow S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)

\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)

\(V\text{ậy}:S< \frac{4}{5}\)

\(-\frac{33}{19}\&-\frac{45}{31}\)

\(-\frac{33}{19}=-\frac{1023}{589}\)(1)

\(-\frac{45}{31}=-\frac{855}{589}\)(2)

Từ (1) và (2) => -33/19>-45/31

Ta có:

31/2.32/2.33/2....60/2=31.32......60/2^30

=(31.32.33....60)(1.2.3....30)/2^30(1.2.3...30)

=(1.3.5...59)(2.4.6...60)/(2.4.6...60)=1.3.5...59

=>P=Q

nhớ ****

cái dòng 3, 4 mk ko hiểu sao 2^30.(1.2.3....30) lại bằng 2.4.6...60

ak hểu r