Đúng rùi bn

BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

a)Ta đặt A=10+15+...+300

Số số hạng của A là:(300-10):5+1=59(số)

Tổng của A là:(10+300).59:2=9145

=>9145+x=6750

=>x=6750-9145

=>x=-2395

b)\(\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}-\frac{1}{x+1}=\frac{59}{77}\)

<=>\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)

<=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)

<=>\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)

<=>\(\frac{6}{7}-\frac{1}{x+1}=\frac{56}{77}\)

<=>\(\frac{1}{x+1}=\frac{6}{7}-\frac{56}{77}=\frac{66}{77}-\frac{56}{77}\)

<=>\(\frac{1}{x+1}=\frac{10}{77}\)

<=>10(x+1)=77

<=>10x+10=77

<=>10x=67

<=>x=6,7

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)\)

\(A=1-\frac{1}{9}=\frac{8}{9}\)

A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

=1\(-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

=1-\(\frac{1}{9}=\frac{8}{9}\)

Vậy A=\(\frac{8}{9}\)

\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(A=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(A=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)

\(A=\frac{9}{10}-\frac{9}{10}=0\)

\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)

\(\Leftrightarrow A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)

\(\Leftrightarrow A=\frac{9}{10}-\frac{9}{10}\)

\(\Leftrightarrow A=0\)

a) 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12  - 1/6 - 1/2

 = 1/10.9 - 1/9.8 - 1/8.7 - 1/7.6 - 1/6.5 - 1/5.4 - 1/4.3 - 1/3.2 - 1/2.1

=  1/10 - 1

=   0,1 - 1

=      -0,9

a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

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

=> \(A=\frac{9}{10}\)

b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)

=> \(A=1+\frac{7}{n-5}\)

Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)

=> n=(-2; 4, 6, 8)

 \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)

\(x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)=\frac{47}{42}\)

\(x+A=\frac{47}{42}\)

     ta thấy :

              \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)

              \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

                  \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

               \(A=\frac{1}{1}-\frac{1}{6}\)

             \(A=\frac{5}{6}\)

       vậy \(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)       

  hay  \(x+\frac{5}{6}=\frac{47}{42}\)

     \(x=\frac{47}{42}-\frac{5}{6}\)

 \(x=\frac{2}{7}\)

\(x+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}=\frac{47}{42}\)

\(x=\frac{47}{42}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)

\(x=\frac{47}{42}-\frac{5}{6}\)

\(x=\frac{2}{7}.\)