A=\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\)

299.A= 299.(\(\frac{1}{1.300}+\frac{1}{2.301}+\frac{1}{3.302}+...+\frac{1}{101.400}\))

299.A=\(\frac{299}{1.300}+\frac{299}{2.301}+\frac{299}{3.302}+...+\frac{299}{101.400}=\frac{1}{1}-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+...+\frac{1}{101}-\frac{1}{400}\)

A= \(=\frac{1}{299}\left(1+\frac{1}{2}+...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)

Tương tự 

B=\(\frac{1}{101}.\left(\frac{1}{1}-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+...+\frac{1}{299}-\frac{1}{400}\right)\)

B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)

B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{400}\right)\)

B= \(\frac{1}{101}.\left(\frac{1}{1}+\frac{1}{2}...+\frac{1}{101}-\frac{1}{300}-\frac{1}{301}-...-\frac{1}{400}\right)\)

Hai dấu ngoặc ở biểu thức A và biểu thức B như nhau

Vậy \(A:B=\frac{1}{299}:\frac{1}{101}=\frac{101}{299}\)

t=(1-1/2+1-1/6+1-1/12+..........+1-1/90+1-1/110)-10/11

t=(1+1+1+...........+1+1)-(1/2+1/6+1/12+.......+1/90+1/110)-10/11

có 10 số 1 vì từ 1/2 đến 109/110 có 10 p/số

t=1.10-(1/1.2+1/2.3+1/3.4+.........+1/9.10+1/10.11)-10/11

t=10-(1-1/2+1/2_1/3+1/3_1/4+.............+1/9-1/10+1/10-1/11)-10/11

t=10-(1-1/11)-10/11

t=10-10/11-10/11

t=10-(10/11+10/11)

t=10-10/11.2

t=10-20/11

t=110/11-20/11

t=90/11

\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)=8+\frac{1}{10}=\frac{81}{10}\)

\(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\)

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

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

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

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

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

\(=8+\frac{1}{10}=\frac{81}{10}\)

\(=8+\frac{1}{10}=\frac{81}{10}\)

\(\frac{81}{10}\)

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

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

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

\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

T = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}+\frac{10}{11}\)

T = \(\frac{10}{11}+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{110}\right)\)

T = \(\frac{10}{11}+\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)

T = \(\frac{10}{11}+10+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)

T = \(\frac{120}{11}+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)

T = \(\frac{120}{11}-\left(1-\frac{1}{11}\right)\)

T = \(\frac{120}{11}-\frac{9}{11}\)

T = \(\frac{111}{11}\)

bài của bạn Giang tính nhầm bước thứ hai từ cuối lên

Sửa: T = \(\frac{120}{11}-\frac{10}{11}=10\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\) \(\frac{89}{90}\)

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

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

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\) 

\(=9-\frac{11}{10}\)

\(=\frac{79}{10}\)

~Học tốt nha~

Đặt : \(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

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

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

\(\Leftrightarrow A=9-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

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

\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{90}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

\(=9-\frac{9}{10}=\frac{81}{10}\)

đề bài đâu mà tính

đề đâu

mày bị ngu dân à! đề đâu