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Ta đặt A=\(-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(\Rightarrow A=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)= \(-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
= - \(\left(1-\dfrac{1}{10}\right)=-\left(\dfrac{10-1}{10}\right)=-\dfrac{9}{10}\)
Ta có: \(-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)\)
\(=-\left(-\dfrac{1}{10}+1\right)\)
\(=-\left(1-\dfrac{1}{10}\right)\)
\(=-\left(\dfrac{10}{10}-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\frac{8}{9}\)
\(=0\)
Chúc bạn học tốt !!!!
Phân tích: 9 phân số thành phần trong tổng trên đều có mẫu số - tử số =1
Do đó, ta tách từng phân số trên thành phân số có hiệu giữa 1 và phân số phần bù của nó.
Từ đó ta thực hiện phép tính dễ dàng.
Đặt tổng trên bằng A ta có:
A = 1 - 1/2 + 1 - 1/6 + 1 - 1/12 + 1 - 1/20 + 1 - 1/30 + 1 - 1/42 + 1 - 1/56 + 1 - 1/72 + 1 - 1/89
= (1 + 1 + 1 + .... 1 + 1) + (1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90)
= 9 - (1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10)
Mỗi phân số thành phần trong biểu thức () ta tách được như sau:
1/1 x 2 = 1- 1/2
1/2 x 3 = 1/2 - 1/3
1/3 x 4 = 1/3 - 1/4
........
1/9 x 10 = 1/9 - 1/10
Như vậy:
A = 9 - (1-1/2 + 1/2-1/3 + 1/3-1/4 + 1/4-1/5 + 1/5-1/6 + 1/6-1/7 + 1/7-1/8 + 1/8-1/9 + 1/9-1/10)
= 9 - (1 - 1/10)
= 9 - 9/10
= 81/10
Đáp số: 81/10
A= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90
=1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5)
+1/(5.6)+1/(6.7)+1/(7.8)
+1/(8.9)+1/(9.10)
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6...
+1/9-1/10
=1-1/10
=9/10
\(\frac{9}{8}-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{72}=\frac{9}{8}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{72}\right)\)
\(=\frac{9}{8}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)\)
\(=\frac{9}{8}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{9}{8}-\left(1-\frac{1}{9}\right)=\frac{9}{8}-\frac{8}{9}=\frac{17}{72}\)
Ta có: -1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= (-1).(1/90 + 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= (-1).(1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90)
= (-1).(1/1 - 1/2 + 1/2 - 1/3 + ... + 1/9 - 1/10)
= (-1).(1 - 1/10)
= (-1).(9/10)
= -9/10
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(A=\frac{9}{10}-\frac{9}{10}=0\)
\(A=\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(\Leftrightarrow A=\frac{9}{10}-\frac{9}{10}\)
\(\Leftrightarrow A=0\)
= - ( 1/2 +1/6+1/12+1/20+ 1/30+ 1/42+ 1/56+ 1/72+ 1/90)
= - ( 1 - 1/2 + 1/2 -1/3 +1/3 -1/4 +1/4 - 1/5 +1/5 -1/6 +1/6 -1/7 +1/7 -1/8 +1/8 -1/9 +1/9 -1/10)
= - ( 1- 1/10 )
= -9/10
\(-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(-\frac{1}{90}\right)+\left(-\frac{1}{72}\right)+\left(-\frac{1}{56}\right)+......+\left(-\frac{1}{2}\right)\)
\(=\left(-1\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{9.10}\right)\)
\(=\left(-1\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...........+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\left(-1\right).\left(1-\frac{1}{90}\right)=\frac{\left(-1\right).89}{90}=-\frac{89}{90}\)
nha
Ta có: \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}=0\)