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a) \(\left(\left(\frac{-12}{16}\right)+\frac{7}{14}\right)-\left(\frac{1}{13}-\frac{3}{13}\right)\) \(=\left(\left(\frac{-3}{4}\right)+\frac{1}{2}\right)-\left(\frac{-2}{13}\right)\) \(=\left(\frac{-2}{8}\right)-\left(\frac{-2}{13}\right)\) \(=\left(\frac{-10}{104}\right)\) \(=\left(\frac{-5}{72}\right)\) | b) \(\frac{10}{11}+\frac{4}{11}:4-\frac{1}{8}\) \(=\frac{10}{11}+\frac{4}{11}:\frac{4}{1}-\frac{1}{8}\) \(=\frac{10}{11}+\frac{4}{11}\cdot\frac{1}{4}-\frac{1}{8}\) \(=\frac{10}{11}+\frac{1}{11}-\frac{1}{8}\) \(=\frac{11}{11}-\frac{1}{8}\) \(=1-\frac{1}{8}\) \(=\frac{7}{8}\) |
HT
a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79
= 1/10 - 1/79
= máy tính ok
mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha
a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)
b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)
c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)
d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)
e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)
g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)
a) 2/9 - (1/20 + 2/9)
= 2/9 - 1/20 - 2/9
= (2/9 - 2/9) - 1/20
= 0 - 1/20
= -1/20
b) -3/14 + 2/13 + (-25/14) + (-15/13)
= (-3/14 - 25/14) + (2/13 - 15/13)
= -2 - 1
= -3
c) -3/11 + 11/8 - 3/8 + (-8/11)
= (-3/11 - 8/11) + (11/8 - 3/8)
= -1 + 1
= 0
d) 3/8 + (-1/4) - (7/12 - 1/6)
= 1/8 - 5/12
= -7/24
e) (1/3 + 12/67 + 13/41) - (79/67 - 28/41)
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41
= 1/3 + (12/67 - 79/67) + (13/41 + 28/41)
= 1/3 - 1 + 1
= 1/3
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
TL :
a, ( -12/16 + 7/14 ) - ( 1/13 - 3/13 )
= ( -3/4 + 1/2 ) - (-2/13)
= (-3/4 + 2/4 ) - ( -2/13 )
= -1/4 - ( -2/13 )
= (-13/52 ) - (-8/52)
= -5/52
b, 10/11 + 4/11 : 4 - 1/8 = 10/11 + 1/11 - 1/8
= 11/11 - 1/8
= 1 -1/8
= 8/8 - 1/8
= 7/8
HT
7/8
HTTTTTTTTTTTTTTTTTTT