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a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
B=1/4x (7+9) + 1/5x(9+11) + 1/6x(11+13) + 1/7x13
=4+2+4+\(\frac{13}{7}\)=10+\(\frac{13}{7}\)=\(\frac{83}{7}\)
a)\(\frac{3}{5}x\frac{7}{9}+\frac{11}{9}x\frac{3}{5}-\frac{3}{10}\)
\(=\frac{3}{5}x\left(\frac{7}{9}+\frac{11}{9}\right)-\frac{3}{10}\)
\(=\frac{3}{5}-\frac{3}{10}\)
\(=\frac{3}{5}\)
tương tự cho phần b nhé
chúc học tốt
=
a.\(\frac{9}{10}\)
b.\(\frac{20}{13}\)
cho mình nha, mình không có thời gian để giải công thức nên mình viết được chừng này thôi ^ ^
a)\(2-3+5-7+9-11+13-15+17=\left(2+5+9+13+17\right)-\left(3+7+11+15\right)\)
\(=46-36=10\)
b)\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{8.9}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{1}-\frac{1}{9}=\frac{9}{9}-\frac{1}{9}=\frac{8}{9}\)
Áp dụng \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Chúc bạn học tốt
Đề là tính bằng cách hợp lý đúng ko bạn
a, 2-3+5-7+9-11+13-15+17
= (5+13) - (3+15) + (2+9-11) + (17-7)
= 18 - 18 + 0 +10
= 10
b, \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
a,3/5*7/9+11/9*3/5-3/10=3/5*(7/9+11/9)-3/10
=3/5*18/9-3/10
=3/5*2-3/10
=6/5-3/10
=12/10-3/10
=9/10
b,14/13:1/2+3/13*2-2*7/13=14/13*2+3/13*2-2*7/13
=2*(14/13+3/13-7/13)
=2*10/13
=20/13
a/ ( 7/9 + 11/9 ) x 3/5 - 3/10
= 2 x 3/5 - 3/10
= 6/5 - 3/10
= 9/10
b/ 14/13 : 1/2 + 3/13 x 2 - 2 x 7/13
14/13 x 2/1 + 3/13 x 2 - 2 x 7/13
2 x ( 14/13 + 3/13 - 7/13 )
2 x 10/13
20/13
A)\(\frac{3}{5}\times\frac{7}{9}+\frac{11}{9}\times\frac{3}{5}-\frac{3}{10}\\ =\frac{3}{5}\times\left(\frac{7}{9}+\frac{11}{9}\right)-\frac{3}{10}\\ =\frac{3}{5}\times\frac{18}{9}-\frac{3}{10}=\frac{3\times2}{5}-\frac{3}{10}\)
\(=\frac{6}{5}-\frac{3}{10}=\frac{12}{10}-\frac{3}{10}=\frac{9}{10}\)
Bài 1:
a) 8 x 4 x 125 x 25
= ( 8 x 125 ) x ( 4 x 25 )
= 1000 x 100
= 100000
b) 2 x 178 x 5
= ( 2 x 5 ) x 178
= 10 x 178
= 1780
Bài 2:
a) 5/7 + 7/13 + 9/13 + 6/5 + 9/7 + 9/5
= ( 5/7 + 9/7 ) + ( 7/13 + 9/13 ) + ( 6/5 + 9/5 )
= 2 + 16/13 + 3
= 81/13
b) 1/11 + 2/11 + ... + 10/11
= 55/11 = 5