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NT
0
TU
1
19 tháng 1 2022
\(\text{15 × 85 : 96 + 55 - 68 : 67 × 152 - 9. }\)
\(=1275:96+46-\dfrac{68}{67}\times152=\dfrac{425}{32}+46-\dfrac{10336}{67}=\dfrac{1897}{32}-\dfrac{10336}{67}=-94,98740672.\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 12 2023
Bài 1:
a, 58.32 + 58.68 - 800
= 58.(32 + 68) - 800
= 58.100 - 800
= 5800 - 800
= 5000
b, 12020 + 280 : [55 - (7 - 4)3]
= 12020 + 280 : [ 55 - 33 ]
= 12020 + 280 : [ 28]
= 12020 + 10
= 12030
c, (96 - 19 - 45) - (55 + 96 - 119)
= 96 - 19 - 45 - 55 - 96 + 119
= (96 - 96) + (119 - 19) - (45 + 55)
= 0 + 100 - 100
= 0
KL
3 tháng 8 2017
1. Ta có : (\(\dfrac{-3}{8}\))3 < 0
(\(\dfrac{8}{243}\))3 > 0
=> (\(\dfrac{-3}{8}\))3 < (\(\dfrac{8}{243}\))3
@Cuber Việt
MV
3 tháng 8 2017
\(\left(\dfrac{-3}{8}\right)^3< 0< \left(\dfrac{8}{243}\right)^3\)
Vậy \(\left(\dfrac{-3}{8}\right)^3< \left(\dfrac{8}{243}\right)^3\)
\(A=\dfrac{34}{7\cdot13}+\dfrac{51}{13\cdot22}+\dfrac{85}{22\cdot37}+\dfrac{68}{37\cdot49}\\ =\dfrac{17}{3}\cdot\dfrac{6}{7\cdot13}+\dfrac{17}{3}\cdot\dfrac{9}{13\cdot22}+\dfrac{17}{3}\cdot\dfrac{15}{22\cdot37}+\dfrac{17}{3}\cdot\dfrac{12}{37\cdot49}\\ =\dfrac{17}{3}\cdot\left(\dfrac{6}{7\cdot13}+\dfrac{9}{13\cdot22}+\dfrac{15}{22\cdot37}+\dfrac{12}{37\cdot49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\dfrac{6}{49}\\ =\dfrac{34}{49}\)
T
28 tháng 4 2019
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(=13\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
S
24 tháng 8 2019
\(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+......-\frac{1}{49}\right)=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right);B=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-.....-\frac{1}{49}\right)=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\Rightarrow\frac{A}{B}=\frac{17}{13}\)
N
24 tháng 8 2019
Có: \(A=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Ttự, ta đc: \(B=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
Vậy \(\frac{A}{B}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
#Walker
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