thà chết đi còn hơn làm cái đống này mất

Bạn tham khảo ở link này nhé :

Câu hỏi của Tăng Minh Châu - Toán lớp 6 | Học trực tuyến

Đặt vế trái của Bất đẳng thức la A

\(A< \frac{1}{8}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}.\)

\(A< \frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{3}{10}< \frac{5}{10}=\frac{1}{2}\)

hhhhhhhhh

Ta thấy: \(\frac{1}{8}< \frac{1}{2}\)

             \(\frac{1}{11}< \frac{1}{2}\)

              \(\frac{1}{12}< \frac{1}{2}\) 

              \(\frac{1}{13}< \frac{1}{2}\)

              \(\frac{1}{41}< \frac{1}{2}\)

               \(\frac{1}{42}< \frac{1}{2}\)

               \(\frac{1}{43}< \frac{1}{2}\)

=> \(\frac{1}{8}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{41}+\frac{1}{42}+\frac{1}{43}< \frac{1}{2}\)

a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

                                                     \(=1+\left(-1\right)\)

                                                     \(=0\)

b) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)

                                                                    \(=1+\left(-1\right)+0,5\)

                                                                    \(=0,5\)

_Học tốt nha_

a, \(\frac{15}{12}\)+ \(\frac{5}{13}\)- \(\frac{3}{12}\)-\(\frac{18}{13}\)

= \(\frac{5}{4}\)+ \(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)

= \(\left(\frac{5}{4}-\frac{1}{4}\right)\)+ \(\left(\frac{5}{13}-\frac{18}{13}\right)\)

= 1 - 1 = 0

b, \(\frac{11}{24}\)- \(\frac{5}{41}\)+ \(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)

= \(\left(\frac{11}{24}+\frac{13}{24}\right)\)- \(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5

= 1 - 1 + 0,5 = 0,5

c,  \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)

=\(\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{5}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{5}{11}\)

= \(\frac{11}{5}.\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)

= \(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)

=  \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)

= 0

d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)

= \(9.\left(0,75-0,25\right)-2\)

= 9. 0,5 - 2 = 2,5

e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)

= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)

= -1 + 1 - \(\frac{1}{2}\)

= \(-\frac{1}{2}\)

a) 15/12 + 5/13- 3/12 - 18/13 = (15/12 - 3/12) + ( 5/13 - 18/13) =  12/12 + -13/13 = 1 + (-1) = 0

b) 11/24 - 5/41 + 13/24 + 0,5  - 36/41  = (11/24 +13/24) - (5/41+36/41)+0,5  =  1 - 1+0,5  =  0,5

c)  ( -3/4 + 2/3) : 5/11 + (-1/4 + 1/3 ) : 5/11 = -3/4  + 2/3 : 5/11 + -1/4 + 1/3 =  =  [( -3/4 + (-1/4) ] + ( 2/3 + 1/3)  : 5/11

                                                                   = -4/4 + 3/3 : 5/11 = -1 + 1  * 11/5

                                                                   = 0 * 11/5 = 0

d)  (-3) ^2 * (3/4 - 0,25) - ( 3 1/2 - 1 1/2) = 9 * (3/4 - 25/100) - ( 7/2 -3/2) = 9 * ( 3/4 - 1/4) - 4/2

                                                               = 9* 1/2 - 2 =  9/ 2 - 2= 5/2

e) 13/25 + 6/41 - 38/25 + 35/41 - 1/2 = ( 13/25 - 38/25) + ( 6/41 + 35/41) - 1/2 = -25/25 + 41/41 - 1/2 = (-1) + 1 - 1/2 = 0 - 1/2 = -1/2

yutyugubhujyikiu

Ta có công thức : \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)

\(\Rightarrow B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+10}\)

           \(=\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+10\right)10}{2}}\)

           \(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{10.11}\)

           \(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

           \(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

           \(=2.\left(\frac{1}{2}-\frac{1}{11}\right)=2.\frac{9}{22}=\frac{9}{11}\)

B=9/11

B=9/11

a) \(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)

\(5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)

\(4A=5A-A=\frac{1}{5}-\frac{1}{5^{2019}}\)

\(A=\frac{1}{20}-\frac{1}{4.5^{2019}}< \frac{1}{20}< \frac{1}{2}\)

b)  Đề có sai không mà đằng cuối lại là \(\frac{1}{4^2}\)lặp lại lần nữa.
c) \(C=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

\(2C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

\(3C=2C+C=1-\frac{1}{64}< 1\)

\(C< \frac{1}{3}\)

d) Xem lại đề nữa đi e, nếu trừ hai vế cho \(\frac{1}{3}\)thì vế trái > 0 > vế phải rồi
e)  \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)(10 số hạng)
                                                    \(=\frac{10}{50}=\frac{1}{5}\)

Tương tự: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{6}\)

\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{7}\)

\(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}>\frac{1}{8}\)

\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{533}{840}>\frac{490}{840}=\frac{7}{12}\)