1. Tính hợp lí :

a) \(\frac{6}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{14}{11}\)

\(=\frac{5}{7}.\left(\frac{6}{11}+\frac{2}{11}-\frac{14}{11}\right)\)

\(=\frac{5}{7}.\frac{-6}{11}=-\frac{30}{77}\)

b) \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)

\(\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}-\frac{1}{3}\right)\)

\(=\frac{1}{3}.\frac{5}{3}=\frac{5}{9}\)

a) Có vẻ đề o đúng lắm . Theo mình o phải là 11/11 mà 1/11

Ta có \(\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>...>\frac{1}{19}>\frac{1}{20}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

hay \(S>\frac{1}{2}\)

b)Ta có 1998 x 1999 + 3997=(2000-2) x 1999 +3997 = 2000 x 1999 - 2 x 1999 +3997 = 1999 x 2000 -3998 +3997 =1999 x 2000 -1

< 1999 x 2000 +2 

=> 1999 x 2000 +2 / 1998 x 1999 +3997 > 1 hay M>1

Thanks you . Mình sẽ kết bạn với cậu nhé

\(\dfrac{1}{4444}< 1,\dfrac{3}{7}< 1,\dfrac{9}{5}>1,\dfrac{7}{3}>1,\dfrac{14}{15}< 1,\dfrac{16}{16}=1,\dfrac{14}{11}>1\)

1/4 < 1

3/7 < 1

9/5 > 1

7/3 > 1

14/15 < 1

16/16 = 1

14/11 >1

\(\frac{1}{4}< 1\) ; \(\frac{3}{7}< 1\) ; \(\frac{9}{5}>1\)

\(\frac{7}{3}>1\) ; \(\frac{14}{15}< 1\) ; \(\frac{16}{16}=1\) ; \(\frac{14}{11}>1\)

1/4  < 1;     3/7 < 1;     9/5 > 1;        14/15 < 1;        16/16 = 1;       14/11 > 1.

TL

1/4 < 1;

3/7 < 1;

9/5 > 1;

7/3 > 1;

14/15 < 1;

16/16 = 1;

14/11 > 1.

HT

Câu 1 ) Sắp xếp theo thứ tự từ bé đến lớn

11 phần 18;7 phần 9;14 phần 15

Câu 2 ) Sắp xếp theo thứ tự từ bé đến lớn

8 phần 9;15 phần 11;9 phần 5

Câu 3 ) So sánh 

3 phần 4 < 5 phần 9

Câu 4 ) So sánh

8 phần 12 < 25 phần 30

Như lời hứa nhớ k nha

Cảm ơn bạn Mỹ Chi nhiều nhé

a) \(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)

      \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

       \(=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

b) Ta có : A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)

                  \(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

                   \(=3.\left(1-\frac{1}{100}\right)\)

                     \(=3.\frac{99}{100}=\frac{297}{100}\)

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

a) 8/15 + 1 = 23/15

b) 3/14 + 11/14 - 2/3 = 1- 2/3 = 1/3

c) 7/15 * (5/3+2/3) = 7/15 * 7/3= 49/45

d) 6/11 + 5/11 - 3/5 = 1 - 3/5 = 2/5

a=23/18

b=1/3

c=49/4

d=2/5