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11^2+22^2=11^2+22^2
22^2+33^2=22^2+33^2
......
88^2+99^2=88^2+99^2
#)Giải :
a)\(9^2\div\left(27^3.81^2\right)=9^2\div\left[\left(3^3\right)^3.\left(9^2\right)^2\right]=9^2\div\left(3^9.9^4\right)\)
Tự lm típ
a,\(9^2:\left(27^3.81^2\right)=3^4:\left(3^9.3^8\right)=3^4:\left(3^{9+8}\right)=3^4:3^{17}=3^{-13}\)
\(\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}=\frac{9.2}{9}=2\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{6^9.2^{10}+6^{10}.2^{10}}=\frac{2^{19}.3^9+3^9.5.2^{18}}{6^9.2^{10}.\left(1+6\right)}=\frac{2^{18}.3^9.\left(2+5\right)}{2^9.3^9.2^{10}.7}=\frac{2^{18}.7}{2^{19}.7}=\frac{1}{2}\)
C2 :
\(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^{2019}\)
\(2\cdot3\cdot3^{x+1}+4\cdot3^{x+1}=10\cdot3^{2019}\)
\(6\cdot3^{x+1}+4\cdot3^{x+1}=10\cdot3^{2019}\)
\(\left(6+4\right)\cdot3^{x+1}=10\cdot3^{2019}\)
\(10\cdot3^{x+1}=10\cdot3^{2019}\)
\(\Rightarrow x+1=2019\)
\(x=2019-1\)
\(x=2018\)
Vậy x = 2018
Chắc sai =))