62. 10 : {780 : [103 - (2.53 + 35 . 14)]}
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NT
9 tháng 10 2023
{50:5-45:5}x7
={10-9}x7
=1x7
=7
d,
=6^2x10:{780:[1000-15,625+35x14]}
=6^2x10:{780:[1000-15,625+490]}
=6^2x10:{780:1474,375}
=6^2x10:0,52903772785
=36x10:0,52903772785
=360:0,52903772785
=608,480769232
khum bt câu d, làm đúng chưa?
chắc sai á
H
14 tháng 7 2021
\(a,\left\{\left[\left(37+13\right):5\right]-45:5\right\}.7\)
\(=\left\{\left[50:5\right]-45:5\right\}.7\)
\(=\left\{10-45:5\right\}.7\)
\(=\left\{10-9\right\}.7\)
\(=1.7=7\)
2 tháng 12 2021
a) {[(37 + 13) : 5] - 45 : 5} . 7
= {[50 : 5] - 45 : 5} . 7
= {10 - 9} . 7
= 1 . 7
= 7
b) 62 . 10 : {780 : [103 - (2 . 53 + 35 . 14)]}
= 36 . 10 : {780 : [1 000 - (2 . 125 + 35 . 14)]}
= 36 . 10 : {780 : [1 000 - (250 + 490)]}
= 36 . 10 : {780 : [1 000 - 740]}
= 36 . 10 : {780 : 260}
= 36 . 10 : 3
= 360 : 3
= 120
18 tháng 9 2023
\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)
\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)
\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)
3 tháng 5 2015
\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)
\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)
\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)
\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)
2 tháng 7 2018
\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(=\frac{3-1}{3}+\frac{15-1}{15}+\frac{35-1}{35}+\frac{63-1}{63}+\frac{99-1}{99}\)
\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)
\(=5+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
\(=5+\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(=5+\frac{5}{11}=\frac{60}{11}\)
MC
1
27 tháng 5 2021
Sửa đề: \(98+99+\dfrac{142}{144}\) \(\rightarrow\dfrac{98}{99}+\dfrac{143}{144}\)
Giải:
\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+\dfrac{62}{63}+\dfrac{98}{99}+\dfrac{143}{144}+\dfrac{194}{195}\)
\(A=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{195}\right)\)
\(A=7-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{195}\right)\)
\(A=7-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\)
\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\right]\)
\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\right]\)
\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\right]\)
\(A=7-\left[\dfrac{1}{2}.\dfrac{14}{15}\right]\)
\(A=7-\dfrac{7}{15}\)
\(A=\dfrac{98}{15}\)
Chúc bạn học tốt!
BT
7
4 tháng 6 2018
A = \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
A = ( 1 - 1/3 ) + ( 1 - 1/15 ) + ( 1 - 1/35 ) + ( 1 - 1/63 ) + ( 1 - 1/99 )
A = ( 1 + 1 + 1 + 1 + 1 ) - ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
A = 5 - \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
A = 5 - ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 )
A = 5 - ( 1 - 1/11 )
A = 5 - 10/11
A = 45/11
AK
4 tháng 6 2018
Dấu \(.\)là dấu nhân
\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)
\(\Rightarrow A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)
\(\Rightarrow A=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(\Rightarrow A=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\Rightarrow A=5-\frac{1}{2}.\frac{10}{11}\)
\(\Rightarrow A=5-\frac{5}{11}\)
\(\Rightarrow A=\frac{55}{11}-\frac{5}{11}\)
\(\Rightarrow A=\frac{50}{11}\)
~ Ủng hộ nhé
\(62\cdot10:\left\{780:\left[103-\left(2\cdot53+35\cdot14\right)\right]\right\}\)
\(=62\cdot10:\left\{780:\left[103-596\right]\right\}\)
\(=620:\dfrac{-780}{493}\)
\(=620\cdot\dfrac{-493}{780}=\dfrac{-15283}{39}\)