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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)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
\(\dfrac{7}{2}+\dfrac{2}{25}=\dfrac{7\times25}{2\times25}+\dfrac{2\times2}{25\times2}=\dfrac{175}{50}+\dfrac{4}{50}=\dfrac{179}{50}\)
\(\dfrac{23}{5}-\dfrac{11}{3}=\dfrac{23\times3}{5\times3}-\dfrac{11\times5}{3\times5}=\dfrac{69}{15}-\dfrac{55}{15}=\dfrac{14}{15}\)
\(\dfrac{3}{7}-\dfrac{1}{14}=\dfrac{3\times2}{7\times2}-\dfrac{1}{14}=\dfrac{6}{14}-\dfrac{1}{14}=\dfrac{5}{14}\)
\(\dfrac{8}{5}:\dfrac{1}{3}=\dfrac{8}{5}\times3=\dfrac{8\times3}{5}=\dfrac{24}{5}\)
\(2:\dfrac{5}{3}=2\times\dfrac{3}{5}=\dfrac{2\times3}{5}=\dfrac{6}{5}\)
\(\dfrac{4}{5}\times\dfrac{7}{15}=\dfrac{4\times7}{5\times15}=\dfrac{28}{75}\)
Lời giải:
\(\frac{7}{2}+\frac{2}{25}=\frac{7\times 25+2\times 2}{2\times 25}=\frac{179}{50}\)
\(\frac{23}{5}-\frac{11}{3}=\frac{23\times 3-5\times 11}{5\times 3}=\frac{14}{15}\)
\(\frac{3}{7}-\frac{1}{14}=\frac{6}{14}-\frac{1}{14}=\frac{6-1}{14}=\frac{5}{14}\)
\(\frac{8}{5}: \frac{1}{3}=\frac{8}{5}\times 3=\frac{24}{5}\)
$2: \frac{5}{3}=\frac{2\times 3}{5}=\frac{6}{5}$
$\frac{4}{5}\times \frac{7}{15}=\frac{28}{75}$
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, \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
a) 1+2+3+...+2018
=[(2018-1):1+1].(2018+1):2
=2018.2019:2
=2037171
b) 4+7+10+13+...+2017
=[(2017-4):3+1].(2017+4):2
=672.2021:2
=679056
c) 1+2-3-4+5+6-7-8+9+10-11-12+13+14
=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+13+14
=-4+(-4)+(-4)+13+14
=-12+27
=15
d) 23 x 75 + 25 x 23 + 180
=23 x (75+25) +180
=23 x 100 +180
=2300+180
2480
chúc ban hoc tốt nha
a) Số số hạng: ( 2018 - 1 ) : 1 + 1 = 2018
Tổng: 2018 x ( 2018 + 1 ) : 2 = 2 037 171
Bài 2. > , < , = ?
5/7 < 4/3 2/5 < 6/10 1/4 = 3/12 27/36 >
2/9 7/6 > 7/9
7/2 = 2/7 15/23 < 1 27/9 > 2 14/15 < 1 51/17 < 4
A)2/3-(1/6+1/8)=2/3-7/24=3/8
B)5/7*14/11*5/7*2/11*5/11*5/7
=5/7*(14/11*2/11*5/11)
=5/7*140/1331=70/1331
C)4/7*(5/4-2/3)*7/8
=4/7*7/12*7/8=1/3*7/8=7/24
Đ)10/1*3+10/3*5+10/5*7+10/7*9+10/9*11
=30/1+50/3+70/5+90/7+110/9
=10+10+10+10+110/9 =40+110/9
=470/9
sao là nhân đúng ko vậy thì
a) 2/3-(1/6+1/8)
=2/3-14/48
=18/48
b)5/7.14/11.5/7.2/11.5/11.5/7
= [ 5/7.5/7.5/7] . [ 2/11.5/11.14/11]
= 125/343 . 140/1331
= 17500/456533 bạn chưa học lũy thừa đúng ko hả con này lấy cả tử cả mẫu nhân với lũy thừa là xong cực nhanh đó
c) 4/7.(5/4-2/3).7/8
=4/7.10/12.7/8
=40/84.7/8
=280/672
d)10/1.3+10/3.5+10/5.7+10/7.9+10/9.11
= 30/1+50/3+70/5+90/7+110/9
=27010/315
Mai tui giải cho hem
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)