A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

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

\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

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

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

\(=1-\frac{9}{10}\)

\(=\frac{1}{10}\)

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

=   \(\frac{1}{1.2}\)+   \(\frac{1}{2.3}\)+    \(\frac{1}{3.4}\)+   \(\frac{1}{4.5}\)+   ......   +   \(\frac{1}{7.8}\)

=   \(1\)\(-\)\(\frac{1}{8}\)

=    \(\frac{7}{8}\)

thiếu bước :v

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

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

\(=\frac{7}{8}\)

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

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

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

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

\(=\frac{7}{8}\)

Đúng rùi bn

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

Biết trả lời nhưng ko biết dấu nhân ở đâu

1/2+1/6+1/12+1/20+1/30+1/42

= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=1-1/7=6/7

Đáp án là : \(\frac{7}{6}\)

Tk nha!!

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

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

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

\(=1-\frac{1}{8}=\frac{7}{8}\)

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

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

=   \(\frac{1}{1}-\frac{1}{8}\)

=  \(\frac{8}{8}-\frac{1}{8}\)

=  \(\frac{7}{8}\)

NHỚ K CHO MK NHA

MK ĐẢM BẢO ĐÚNG 100% LUÔN

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

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

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+.....+\frac{1}{2}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+........+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{8.9}\right)\)

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

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)

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

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

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

\(=\frac{8}{9}-\frac{8}{9}=0\)

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}...-\frac{1}{6}-\frac{1}{2}\)

= \(\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)

= \(\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)

= \(\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

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

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

= \(\frac{8}{9}-\frac{8}{9}\)

= \(0\)

"girl cute" là sai rồi bạn ơi, trong tiếng anh, tính từ (cute) phải đứng trước danh tư (girl).

0

Chúc bạn học tốt!!
 

từ đầu bài ta có

1 phần 6 cộng một phần12 cộng..... cộng một phần 42

=1phần 2.3 +1 phần 3.4 +....+1 phần6.7

=1 phần2 -(1 phần 3 +1 phần 3)...........-(1 phần 6 +1 phần 6)-1 phần 7

1 phần 2 -1 phần 7

=5 phần 14 

mongcasc bạn giúp mk mai mk ktra chương, tất nhiên mk sẽ k cho ai nhanh mà đúng

Ta có:

\(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(\Rightarrow A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(A=\frac{1}{2}-\frac{1}{7}\)

\(A=\frac{5}{14}\)