\(\frac{12}{7}>1\)

\(\frac{89}{1112}< 1\)

\(\Rightarrow\frac{12}{7}>\frac{89}{1112}\)  ( So sánh với 1)

Các so sánh ta có là so sánh tất cả các phân số với 1.

Ta có: \(\frac{12}{7}>1;\frac{89}{1112}< 1\) nên ta có \(\frac{12}{7}>\frac{89}{1112}\).

Vậy \(\frac{12}{7}>\frac{89}{1112}\).

~ Hok tốt ~

a) Ta có : \(\frac{12}{48}< \frac{12}{47}\); \(\frac{12}{48}< \frac{13}{48}\)

=> \(\frac{12}{48}< \frac{13}{47}\)

b) Ta có : \(\frac{7}{13}=1-\frac{6}{13}\)

               \(\frac{17}{23}=1-\frac{6}{23}\)

Mà \(-\frac{6}{13}< -\frac{6}{23}\)=> \(\frac{7}{13}< \frac{17}{23}\)

a) 3 phân số đó là: 51/80    ,    52/80 và 53/80

bai 2 ket qua la 23

Bài 1:

Ta có:

\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)

Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)

                                                     \(\Leftrightarrow N< M\)

Vậy \(M>N.\)

Bài 2:

Ta có:

\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)

\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)

Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)

\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)

                                                                     \(\Leftrightarrow A>B\)

Vậy \(A>B.\)

Bài 3:

\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)

                                                                \(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)

                                                                \(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)

Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)

\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm

\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)

Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)

Bài 4:

\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)

Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)

\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)

\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)

Vậy \(\frac{1991.1999}{1995.1995}< 1.\)

bài 1

a,

32 + 68 :17 x 5 - 29

= 32 + 20 -29

= 52 - 29

= 23

b,

15 x 48 - 30 x 24 - 125

= 720 - 720 -125

= 0-125

a,

32 + 68 :17 x 5 - 29

= 32 + 20 -29

= 52 - 29

= 23

b,

15 x 48 - 30 x 24 - 125

= 720 - 720 -125

= 0-125

\(vì\hept{\begin{cases}-18>-23\\91< 144\end{cases}}\Rightarrow\frac{-18}{91}>\frac{-23}{144}\)

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ta có

\(1-\frac{2018}{2019}=\frac{1}{2019}\)và\(1-\frac{2019}{2020}=\frac{1}{2020}\)

vì\(\frac{1}{2019}>\frac{1}{2020}\)vậy\(\frac{2018}{2019}>\frac{2019}{2020}\)

a) Ta có \(\frac{13}{7}=2-\frac{1}{7}\)

              \(\frac{21}{12}=2-\frac{1}{4}\)

Vì \(\frac{1}{7}< \frac{1}{4}\)\(\Rightarrow2-\frac{1}{7}>2-\frac{1}{4}\)\(\Rightarrow\frac{13}{7}>\frac{21}{12}\)

Vậy \(\frac{13}{7}>\frac{21}{12}\)

b) Ta có : \(\frac{2018}{2019}=1-\frac{1}{2019}\)

               \(\frac{2019}{2020}=1-\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\Rightarrow1-\frac{1}{2019}< 1-\frac{1}{2020}\Rightarrow\frac{2018}{2019}< \frac{2019}{2020}\)

Vậy \(\frac{2018}{2019}< \frac{2019}{2020}\)

c) Ta có :Vì  \(\frac{17}{53}< \frac{17}{50}< \frac{19}{50}\) \(\Rightarrow\frac{17}{53}< \frac{19}{50}\)

Vậy \(\frac{17}{53}< \frac{19}{50}\)

Giải

Bài 1:

Các số tự nhiên mà chữ số hàng đơn vị hơn chữ số hàng chục 2 đơn vị gồm :

13;24;35;46;57;68;79.

Vậy có tất cả 7 số hạng như thế !

Bài 2:

Hai số tự nhiên giống nhau mà chia 5 dư 3 là 88.

Bài 3:

a)Số lượng số hạng của tổng trên là:

    (403-31):4+1=94(số hạng)

Tổng trên là:

   (403+31).94:2=20 398

Bài 4:

A.4 1/5.10/11+5 2/11

=21/5.10/11+57/11

=42/11+57/11

=99/11

=9

B.1,25+7/8:14/24-1/2

=125/100+7/8:14/24-1/2

=5/4+7/8:7/12-1/2

=5/4+3/2-1/2

=11/4-1/2

=9/4

Tiếp nha !

C.17:17/12-5/12+0,1

=17/1.12/17-5/12+1/10

=12-5/12+1/10

=đề đúng thì phải như thế này

=17/12:17-5/12+0,1

=1/12-5/12+1/10

=-1/3+1/10

=-10/30+3/30

=-7/30.

\(B=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{18\cdot20}\)

\(B=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\)

\(B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\)

\(B=\frac{1}{2}-\frac{1}{20}\)

\(B=\frac{9}{20}\)

=))

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\)

\(A=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)