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1 phần 2 nhân 3 phần 4 cộng 1 phần 2 nhân 1 phần 4 ; 11 phần 3 nhân 26 phần 7 trừ 20 phần 7 nhân 8 phần 30
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`#3107.101107`
\(3^9\div3^8+5^8\div5^7\\ =3^{9-8}+5^{8-7}\\ =3+5\\ =8\)
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\(6^9\div6^7-30\\ =6^{9-7}-30\\ =6^2-30\\ =36-30\\ =6\)
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\(\left(2^7\times2^2\right)\div2^9-5^0\\ =2^{7+2}\div2^9-1\\ =2^9\div2^9-1\\ =2^0-1\\ =1-1\\ =0\)
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\(3\times3\times3\times3\times3\times3^5\\ =3^{1+1+1+1+1+5}\\ =3^{10}\)
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(12x - 43) . 83 = 4 . 84
(12x - 64) . 512 = 4 . 4096
(12x - 64) . 512 = 16 384
(12x - 64) = 16 384 : 512
(12x - 64) = 32
12x = 32 + 64
12x = 96
x = 96 : 12
x = 8
(12x - 43) . 83 = 4 . 84
(12x - 64) . 512 = 4 . 4096
(12x - 64) . 512 = 16 384
(12x - 64) = 16 384 : 512
(12x - 64) = 32
12x = 32 + 64
12x = 96
x = 96 : 12
x = 8
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\(a,\)\(\frac{2^5\times3^{12}\times7^8}{2^7\times3^{10}\times7^9}=\frac{3^2\times\left(2^5\times3^{10}\times7^8\right)}{2^2\times7\times\left(2^5\times3^{10}\times7^8\right)}\)\(=\frac{3^2}{2^2\times7}=\frac{9}{28}\)
\(b,\)Tương tự
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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{6^5\cdot27^2}{7^3\cdot9^5}=\frac{2^5\cdot3^5\cdot\left(3^3\right)^2}{7^3\cdot\left(3^2\right)^5}=\frac{2^5\cdot3^5\cdot3^6}{7^3\cdot3^{10}}=\frac{2^5\cdot3^{11}}{7^3\cdot3^{10}}=\frac{2^5\cdot3}{7^3}\)
b, \(\frac{12^7\cdot9^3}{8^5\cdot27^3}=\frac{3^7\cdot2^{12}\cdot3^6}{2^{15}\cdot3^9}=\frac{2^{12}\cdot3^{13}}{2^{15}\cdot3^9}=\frac{3^4}{2^3}\)
c, \(\frac{20^6\cdot8^2}{16^3\cdot25^3}=\frac{2^{12}\cdot5^6\cdot2^6}{2^{12}\cdot5^6}=2^6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=362880-40320-5040x8^2
=317520x8^2
=2540160^2
=6452412825600
Tk plz
:))
Thanks
=362880-40320-5040x8^2
=317520x8^2
=2540160^2
=6452412825600