\(3\div4=?\)
\(10000\times9-\left(2009-467\right)=?\)
\(98745-4524\)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 4 2023
(\(x\) - 12) : 4 = 7
\(x\) - 12 = 7 \(\times\) 4
\(x\) - 12 = 28
\(x\) = 28 + 12
\(x\) = 40
\(x\) - 12 : 4 = 7
\(x\) - 3 = 7
\(x\) = 7 + 3
\(x\) = 10
26 tháng 4 2023
\(\left(x-12\right)\) : 4 = 7
<=> \(x-12\) = 7 x 4
<=> \(x-12\) = 28
<=> \(x\) = 40
BT
0
TN
15 tháng 7 2016
(5*411-3*165):410
=[5*(22)11-3*(24)5)]:(22)10
=[5*222-3*220]:220
=[5*22*220-3*220]:220
=[20*220-3*220]:220
=220[20-3]:220
=17
NN
16 tháng 2 2023
\(C=\dfrac{5\times2^{12}\times3^8-3^9\times2^{12}}{2^2\times2^{13}\times3^8+2\times2^{12}\times\left(-3^9\right)}=\dfrac{3^8\times2^{12}\times\left(5-3\right)}{2^{15}\times3^8+2^{13}\times\left(-3\right)^9}\)
\(=\dfrac{3^8\times2^{12}\times2}{2^{13}\times3^8\times\left(4-3\right)}=\dfrac{1}{1}=1\)
\(#PaooNqoccc\)
CC
1
4 tháng 4 2017
\(\left(6\frac{5}{5}-2\frac{4}{5}\right).\frac{25}{8}-\frac{8}{5}.\frac{1}{4}\)
\(=\frac{21}{5}.\frac{25}{8}-\frac{2}{5}\)
\(=\frac{105}{8}-\frac{2}{5}=\frac{525}{40}-\frac{16}{40}=\frac{509}{40}\)
3 tháng 9 2018
\(30-\left[4\left(x-2\right)+15\right]=3\) \(\left(8x-120:4\right).3^3=3^6\)
\(4\left(x-2\right)+15=27\) \(\left(8x-120:4\right)=27\)
\(4\left(x-2\right)=12\) \(8x-30=27\)
\(x-2=3\) \(8x=57\)
\(x=5\) \(x=\frac{57}{8}\)
NN
21 tháng 8 2020
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
8 tháng 8 2016
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)
\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)
\(\frac{1}{2x+3}=\frac{1}{48}\)
=> 2x + 3 = 48
=> 2x = 48 - 3
=> 2x = 45
=> x = 45/2
22 tháng 5 2022
\(A=\dfrac{2^{19}\cdot3^9-3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot2^{10}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{19}\cdot3^9-2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\left(2-5\right)}{2^{19}\cdot3^9\cdot7}=\dfrac{1}{2}\cdot\dfrac{-3}{7}=\dfrac{-3}{14}\)
`3:4=3/4`
`10000 x 9 - (2009 - 467)`
`=10000 xx 9 - 1542`
`=90000 - 1542`
`=88458`
`98745 - 4524 = 94221`
1. 3/4
2.88458
3.94221