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B = \(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
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C
1
10 tháng 1 2022
\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\cdot9}=2\)
PH
0
HH
1
C
8 tháng 1
\(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\dfrac{\left(3.2^2.2^{16}\right)^2}{11.2^2.\left(2^2\right).11-\left(2^4\right)^9}\)
\(=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^2.2^{22}-2^{36}}\)
\(=\dfrac{3^2.2^4.2^{32}}{11.2^{24}-2^{36}}\)
\(=\dfrac{3^2.2^{34}}{11.2^{24}-2^{36}}\)
\(=\dfrac{3^2.2^{24}.2^{10}}{11.2^{24}-2^{12}.2^{24}}\)
\(=\dfrac{3^2.2^{24}.2^{10}}{\left(11-2^{12}\right).2^{24}}\)
\(=\dfrac{3^2.2^{10}}{11-2^{12}}\)
QT
3
NT
14 tháng 3 2018
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{35}.2}\)
\(=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{9.2^{35}.2}{2^{35}.9}\)
\(=2\)
HN
14 tháng 3 2018
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
= \(\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
= \(\frac{3^2.4^2.2^{32}}{11.2^{35}-2^{35}.2}\)
= \(\frac{3^2.2^4.2^{32}}{2^{35}.\left(11-2\right)}\)
= \(\frac{3^2.2^{36}}{2^{35}.9}\)
= \(\frac{9.2^{35}.2}{2^{35}.9}\)
= \(2\)
1 tháng 2 2019
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.
KS
1
7 tháng 7 2022
\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^3\cdot2^{22}-2^{36}}\)
\(=\dfrac{3^2\cdot2^{36}}{2^{25}\left(11-2^{11}\right)}=\dfrac{3^2\cdot2^{11}}{11-2048}=\dfrac{-3\cdot2^{11}}{679}\)
27 tháng 2 2023
\(\dfrac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\dfrac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^2\cdot\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\dfrac{3^2\cdot\left(2^2\right)^2\cdot\left(2^{16}\right)^2}{11\cdot2^2\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^4\cdot2^{32}}{11\cdot2^{24}-2^{36}}\)
\(=\dfrac{3^2\cdot2^{34}}{11\cdot2^{24}-2^{36}}=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{11\cdot2^{24}-2^{12}\cdot2^{24}}\)
\(=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{\left(11-2^{12}\right)\cdot2^{24}}=\dfrac{3^2\cdot2^{10}}{11-2^{12}}\)
\(B=\dfrac{3^2.4^2.2^{18}}{11.2^{13}.2^{22}-2^{36}}\)
\(B=\dfrac{3^2.2^{22}}{2^{35}\left(11-2\right)}\)
\(B=\dfrac{3^2}{2^{13}.9}\)
\(B=\dfrac{1}{2^{13}}=\dfrac{1}{8129}\)