A=4/2.10+4/10.18+4/18+26+.......+4/202+210
Tính (A.23-25.43)/(45.34)
K
Khách
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P
13 tháng 7 2023
a) \(227+50+23=\left(227+23\right)+50=250+50=300\)
b) \(135+360+65+40=\left(135+65\right)+\left(360+40\right)=200+400=600\)
c) \(1+2+3+4+5+...+97+98+99+100\)
\(=\left(100+1\right)+\left(99+2\right)+...+\left(50+51\right)\)
\(=101+101+101+...+101\)
\(=101\cdot50\)
\(\Leftrightarrow5050\)
d) \(115\cdot13-13\cdot15=13\cdot\left(115-15\right)=13\cdot100=1300\)
e) \(50-49+48-47+...+4-3+2-1\)
\(=\left(50-49\right)+\left(48-47\right)+...+\left(2-1\right)\)
\(=1+1+1+1+..+1\)
\(=1\cdot25\)
\(=25\)
P
13 tháng 7 2023
f) \(30\cdot40\cdot50\cdot60=10\cdot3+10\cdot4+10\cdot5+10\cdot6\)
\(=10\cdot10\cdot10\cdot10\cdot3\cdot4\cdot5\cdot6\)
\(=10000\cdot360\)
\(=3600000\)
g) \(27\cdot36+27\cdot64=27\cdot\left(36+64\right)=27\cdot100=2700\)
h) \(5\cdot2^2-18:3=5\cdot4-18:3=20-6=14\)
i) \(13\cdot17-256:16+14:7-2021^0\)
\(=13\cdot17-4^4:4^2+2-1\)
\(=13\cdot17-16+2-1\)
\(=13\cdot17-17\)
\(=17\cdot\left(13-1\right)\)
\(=204\)
j) \(7^2-36:3=49-12=37\)
T
3 tháng 8 2023
a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
NT
Tính nhanh
19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 25 + 26
1/3 + 1/4 + 1/5 + 4/6 + 9/12 + 16/20
1
27 tháng 4 2023
\(19+18+17+16+14+21+22+23+24+25+26\)
\(=\left(19+21\right)+\left(18+22\right)+\left(17+23\right)+\left(16+24\right)+\left(14+26\right)+25\)
\(=30+30+30+30+30+25\)
\(=175\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{6}+\dfrac{9}{12}+\dfrac{16}{20}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\)
\(\text{=}1+1+1\)
\(\text{=}3\)
HN
1
25 tháng 10 2017
\(A=1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+\dfrac{26}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{26}{21}\right)\)
\(=1+\dfrac{31}{21}\)
\(=1\dfrac{31}{21}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 12 2023
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
NK
1
13 tháng 7 2017
\(\dfrac{7}{2.10}+\dfrac{7}{10.18}+...+\dfrac{7}{82.90}=\dfrac{7}{8}\left(\dfrac{8}{2.10}+\dfrac{8}{10.18}+...+\dfrac{8}{82.90}\right)=\dfrac{7}{8}\left(\dfrac{1}{2}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{18}+...+\dfrac{1}{82}-\dfrac{1}{90}\right)=\dfrac{7}{8}\left(\dfrac{1}{2}-\dfrac{1}{90}\right)=\dfrac{7}{8}.\dfrac{22}{45}=\dfrac{77}{180}\)
Nhận xét:
\(\frac{4}{n \left(\right. n - 1 \left.\right)} = 4 \left(\right. \frac{1}{n - 1} - \frac{1}{n} \left.\right)\)
Nên tổng sẽ rút gọn dạng dây chuyền:
\(A=4\left(\right.1-\frac{1}{2}+\frac{1}{9}-\frac{1}{10}+.\ldots\ldots+\frac{1}{201}-\frac{1}{202}\left.\right)\)
Các cặp triệt tiêu dần, còn lại:
\(A = 4 \left(\right. 1 - \frac{1}{202} \left.\right) = 4 \cdot \frac{201}{202} = \frac{804}{202} = \frac{402}{101}\)
Tính tiếp:
\(0,23 - 25,43 = - 25,2 = - \frac{252}{10}\) \(45,34 = \frac{4534}{100}\)
Giá trị cần tìm:
\(\frac{A \left(\right. 0,23 - 25,43 \left.\right)}{45,34} = \frac{\frac{402}{101} \cdot \left(\right. - \frac{252}{10} \left.\right)}{\frac{4534}{100}}\)
Rút gọn:
\(= \frac{402 \cdot \left(\right. - 252 \left.\right) \cdot 100}{101 \cdot 10 \cdot 4534} = \frac{- 10130400}{458 , ?}\)
Rút gọn hết:
\(= - \frac{2520}{11335}\)
Kết quả cuối cùng:
\(\boxed{- \frac{2520}{11335}}\)
Tiểu đệ xin cáo lui!
Ta có: \(A=\frac{4}{2\cdot10}+\frac{4}{10\cdot18}+\cdots+\frac{4}{202\cdot210}\)
\(=\frac12\left(\frac{8}{2\cdot10}+\frac{8}{10\cdot18}+\cdots+\frac{8}{202\cdot210}\right)\)
\(=\frac12\left(\frac12-\frac{1}{10}+\frac{1}{10}-\frac{1}{18}+\cdots+\frac{1}{202}-\frac{1}{210}\right)=\frac12\left(\frac12-\frac{1}{210}\right)\)
\(=\frac12\cdot\frac{104}{210}=\frac{52}{210}=\frac{26}{105}\)