6 So as to get the correct word meaning, you should look it up in the dictionary

7 Peter has tried very hard to achieve his goal

8 I was exhausted from sitting for 4 exams in a row

mẫu số chung của các số là 2520.                                                          1/3 sẽ được 840/2520                                                                            1/6 sẽ đc 420/2520                                                                               1/10 sẽ đc 252/2520                                                                           1/21 sẽ đc 120/2520                                                                              1/28 sẽ đc 90/2520                                                                                 1/36 sẽ đc 70/2520                                                                                  1/45 sẽ đc 56/2520                                                                               cộng các tử số ta được:                                                                        840+420+252+120+90+70+56/2520= 1848/2520                                  rút gọn ta được :11/15

Mẫu số chung: 20

\(\dfrac{1}{2}=\dfrac{10}{20};\dfrac{2}{5}=\dfrac{8}{20};\dfrac{11}{20};\dfrac{3}{10}=\dfrac{6}{20}\)

Vậy kéo co được nhiều số học sinh đăng kí nhất, cầu lông có số học sinh đăng kí ít nhất.

cảm ơn bn nguyễn đăng nhân nha

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{16}=0\)

\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{16}\)

\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{4}\\x-\dfrac{1}{3}=-\dfrac{1}{4}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)

a)

Mẫu số chung 2 phân số: 63

\(\dfrac{-3}{7}=\dfrac{-3\cdot9}{7\cdot9}=\dfrac{-27}{63}\)

\(\dfrac{-4}{9}=\dfrac{-4\cdot7}{9\cdot7}=\dfrac{-28}{63}\)

Vì \(-27>-28\) nên\(\dfrac{-27}{63}>\dfrac{-28}{63}\)

Vậy \(\dfrac{-3}{7}>\dfrac{-4}{9}\)

b)

\(\dfrac{10}{15}=\dfrac{10:5}{15:5}=\dfrac{2}{3}\)

\(\dfrac{12}{16}=\dfrac{12:4}{16:4}=\dfrac{3}{4}\)

\(\dfrac{2}{3}=1-\dfrac{1}{3}\)

\(\dfrac{3}{4}=1-\dfrac{1}{4}\)

\(\Rightarrow\dfrac{3}{4}>\dfrac{2}{3}\)

c) 

\(\dfrac{99}{-98}< 0< \dfrac{33}{49}\Rightarrow\dfrac{99}{98}< \dfrac{33}{49}\)

d) \(\dfrac{105}{106}< 1< \dfrac{94}{93}\Rightarrow\dfrac{105}{106}< \dfrac{94}{93}\)

e) 

\(\dfrac{63}{64}=1-\dfrac{1}{64}\)

\(\dfrac{32}{33}=1-\dfrac{1}{33}\)

Vì \(64>33\) nên \(\dfrac{1}{64}< \dfrac{1}{33}\)

\(\Rightarrow1-\dfrac{1}{64}>1-\dfrac{1}{33}\)

Vậy \(\dfrac{63}{64}>\dfrac{32}{33}\)

f)

\(\dfrac{2020}{2019}=1+\dfrac{1}{2019}\)

\(\dfrac{2022}{2021}=1+\dfrac{1}{2021}\)

Vì \(2019< 2021\) nên \(\dfrac{2020}{2019}>\dfrac{2022}{2021}\)

\(\Rightarrow1+\dfrac{1}{2019}>1+\dfrac{1}{2021}\)

Vậy \(\dfrac{2020}{2019}>\dfrac{2022}{2021}\)

         A =   \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

       3A =   1   -  \(\dfrac{2}{3^{ }}\) +   \(\dfrac{3}{3^2}\) - \(\dfrac{4}{3^3}\) + ... + \(\dfrac{99}{3^{98}}\) - \(\dfrac{100}{3^{99}}\)

3A+A = 1-\(\dfrac{2}{3^{ }}\)+\(\dfrac{3}{3^2}\)-\(\dfrac{4}{3^3}\)+...+\(\dfrac{99}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)+\(\dfrac{1}{3}-\dfrac{2}{3^2}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

4A = 1-(\(\dfrac{2}{3}\)-\(\dfrac{1}{3}\)) +(\(\dfrac{3}{3^2}\)-\(\dfrac{2}{3^2}\))-(\(\dfrac{4}{3^3}\)-\(\dfrac{3}{3^3}\))+...+(\(\dfrac{99}{3^{98}}\)-\(\dfrac{98}{3^{98}}\))-(\(\dfrac{100}{3^{99}}\)-\(\dfrac{99}{3^{99}}\))-\(\dfrac{100}{3^{100}}\)

   4A = 1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{98}}\)-\(\dfrac{1}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)

12A =  3-1+\(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)+....+\(\dfrac{1}{3^{97}}\)-\(\dfrac{1}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)

12A+4A=3-1+\(\dfrac{1}{3}\)-\(\dfrac{1}{3^2}\)+..+\(\dfrac{1}{3^{97}}\)-\(\dfrac{1}{3^{98}}\)-\(\dfrac{100}{3^{99}}\)+1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3^2}\)-\(\dfrac{1}{3^3}\)+..+\(\dfrac{1}{3^{98}}\)-\(\dfrac{1}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)

16A = 3+(-1+1)+(\(\dfrac{1}{3}-\dfrac{1}{3}\))+...+(-\(\dfrac{1}{3^{98}}\)+\(\dfrac{1}{3^{98}}\))+(-\(\dfrac{100}{3^{99}}\)-\(\dfrac{1}{3^{99}}\)) - \(\dfrac{100}{3^{100}}\)

16A = 3 - \(\dfrac{101}{3^{99}}\)  - \(\dfrac{100}{3^{100}}\)

16A = 3 - \(\dfrac{303}{3^{100}}\) - \(\dfrac{100}{3^{100}}\)

16A = 3 - \(\dfrac{403}{3^{100}}\)

A = \(\dfrac{3}{16}\) - \(\dfrac{403}{16.3^{100}}\) < \(\dfrac{3}{16}\) < \(\dfrac{3}{14}\) (đpcm)

Đoạn trích trên sử dụng bptt nhân hóa

được sử dụng với tác dụng giúp cho các sự vật đều trở nên sống động và gần gũi hơn với con người

 

\(a,\dfrac{7}{-9}+\dfrac{-1}{-9}=\dfrac{-7}{9}+\dfrac{1}{9}=\dfrac{-7+1}{9}=\dfrac{-6}{9}=\dfrac{-2}{3}\\ b,\dfrac{7}{-18}+\left(\dfrac{-5}{12}-\dfrac{13}{-18}\right)=\dfrac{-7}{18}-\dfrac{5}{12}+\dfrac{13}{18}=\left(\dfrac{13}{18}-\dfrac{7}{18}\right)-\dfrac{5}{12}\\ =\dfrac{6}{18}-\dfrac{5}{12}=\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{1.4-5}{12}=\dfrac{-1}{12}\\ c,5-\dfrac{-7}{8}+\dfrac{15}{-20}=5+\dfrac{7}{8}-\dfrac{3}{4}=\dfrac{5.8+7-3.2}{8}=\dfrac{40+7-6}{8}=\dfrac{41}{8}\)

a) \(\dfrac{7}{-9}+\dfrac{-1}{-9}=\dfrac{6}{-9}=\dfrac{-2}{3}\)

b) \(\dfrac{7}{-18}+\left(\dfrac{-5}{12}-\dfrac{13}{-18}\right)\)

\(=\dfrac{7}{-18}-\dfrac{5}{12}-\dfrac{13}{-18}\)

\(=\dfrac{-6}{-18}-\dfrac{5}{12}\)

\(=\dfrac{1}{3}-\dfrac{5}{12}=\dfrac{4}{12}-\dfrac{5}{12}\)

\(=\dfrac{-1}{12}\)

c) \(5-\dfrac{-7}{8}+\dfrac{15}{20}\)

\(=5-\dfrac{-7}{8}+\dfrac{3}{4}\)

\(=5-\dfrac{-7}{8}+\dfrac{6}{8}\)

\(=5-\dfrac{-1}{8}=5+\dfrac{1}{8}\)

\(=\dfrac{41}{8}\)

1; (\(\dfrac{-4}{25}\)).(-\(\dfrac{-25}{8}\))

 = \(\dfrac{-4.25}{25.4.2}\)

= \(-\dfrac{1}{2}\)

2; \(\dfrac{5}{-14}\).(\(\dfrac{-7}{10}\))

= \(\dfrac{5.\left(-7\right)}{2.5.\left(-7\right).2}\) 

= \(\dfrac{1}{4}\)

3; \(\dfrac{-15}{4}\).(\(\dfrac{-16}{25}\))

= \(\dfrac{3.5.4.4}{4.5.5}\)

= \(\dfrac{12}{5}\)

4; 15. (- \(\dfrac{13}{10}\))

 = 5.3.\(\dfrac{\left(-13\right)}{2.5}\)

= - \(\dfrac{39}{2}\)

5koϵD ;7ϵD ;0koϵD  ;10ϵD

\(D=\left\{6;7;8;9;10;11\right\}\)

\(D=\left\{x|x\in N,5< x< 12\right\}\)

____

\(5\notin D\)

\(7\in D\)

\(17\notin D\)

\(0\notin D\)

\(10\in D\)