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ND
0
NL
0
TM
1
AH
Akai Haruma
Giáo viên
26 tháng 1
Lời giải:
$A=\frac{15-5}{5.15}+\frac{31-15}{15.31}+\frac{45-31}{31.45}+\frac{52-45}{45.52}+\frac{65-52}{52.65}+\frac{1}{13.70}+\frac{1}{70.15}$
$=\frac{1}{5}-\frac{1}{15}+\frac{1}{15}-\frac{1}{31}+\frac{1}{31}-\frac{1}{45}+\frac{1}{45}-\frac{1}{52}+\frac{1}{52}-\frac{1}{65}+\frac{1}{70}(\frac{1}{13}+\frac{1}{15})$
$=\frac{1}{5}-\frac{1}{65}+\frac{1}{70}.\frac{28}{195}$
$=\frac{12}{65}+\frac{2}{95}$
$=\frac{254}{1325}$
25 tháng 6 2018
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
18 tháng 2 2021
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PK
4
HV
28 tháng 8 2021
\(=\dfrac{\left(2^4\right)^5.\left(3.5\right)^5}{2^{10}.3^5.5^4}=\dfrac{2^{20}.3^5.5^5}{2^{10}.3^5.5^4}=2^{10}.5=1024.5=5120\)
tik mik nha
LN
0
2 tháng 5 2019
a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)
ta có :
\(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)
\(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)
\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)
Vậy \(A< 3\)
KN
2 tháng 5 2019
a. Ta có :
\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)
\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)
\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)
Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)
Vậy \(A< 3\)
H
10 tháng 1 2023
`a, 2/3 + 1/5 . 10/7`
`= 2/3 + 10/35`
`=2/3 +2/7`
`=14/21 + 6/21`
`=20/21`
`b, 2/7 + 5/7 . 14/25`
`= 2/7 +2/5`
`= 10/35+14/35`
`= 24/35`
`c, 7/12 - 27/7 . 1/18`
`= 7/12 - 3/14`
`= 31/84`
`d, 3/10 . (-5/6) - 1/8`
`= -1/4 -1/8`
`= -2/8 -1/8`
`= -3/8`
VN
10 tháng 1 2023
a)
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
b)
\(\dfrac{2}{7}+\dfrac{5}{7}.\dfrac{14}{25}\)
\(=\dfrac{2}{7}+\dfrac{2}{5}\)
\(=\dfrac{10}{35}+\dfrac{14}{35}\)
\(=\dfrac{24}{35}\)
c)
\(\dfrac{7}{12}-\dfrac{27}{7}.\dfrac{1}{18}\)
\(=\dfrac{7}{12}-\dfrac{3}{14}\)
\(=\dfrac{98}{168}-\dfrac{36}{168}\)
\(=\dfrac{31}{84}\)
d)
\(\dfrac{3}{10}.\left(-\dfrac{5}{6}\right)-\dfrac{1}{8}\)
\(=-\dfrac{1}{4}-\dfrac{1}{8}\)
\(=-\dfrac{2}{8}-\dfrac{1}{8}\)
\(=\dfrac{-3}{8}\)
O
10 tháng 3 2018
\(a.-\dfrac{1}{5}.\dfrac{1}{2}\)
\(=\dfrac{-1.1}{5.2}\)
\(=\dfrac{-1}{10}.\)
\(b.-\dfrac{1}{8}.\dfrac{8}{9}\)
\(=\dfrac{-1.8}{8.9}\)
\(=\dfrac{-1}{9}.\)
\(c.-\dfrac{3}{7}.\dfrac{14}{15}\)
\(=\dfrac{-3.14}{7.15}\)
\(=\dfrac{-3.7.2}{7.3.5}\)
\(=\dfrac{-2}{5}.\)
\(d.-\dfrac{7}{5}.\dfrac{15}{21}\)
\(=\dfrac{-7.15}{5.21}\)
\(=\dfrac{-7.5.3}{5.7.3}\)
\(=-1.\)
\(e.-4.\dfrac{7}{24}\)
\(=\dfrac{-4.7}{24}\)
\(=\dfrac{-4.7}{4.6}\)
\(=\dfrac{-7}{6}.\)
\(f.-\dfrac{9}{13}.\dfrac{5}{18}\)
\(=\dfrac{-9.5}{13.18}\)
\(=\dfrac{-9.5}{13.9.2}\)
\(=\dfrac{-1.5}{13.2}\)
\(=\dfrac{-5}{26}.\)
=6/7-[(10/7:21/5)-1]
=6/7-[50/147-1]
=6/7-(-97/147)
=6/7+97/147
=1561/1029=>TỐI GIẢN =223/147 NHA