`1`
`a)|x+2010|=2011`
`→` \(\left[ \begin{array}{l}x+2010=2011\\x+2010=-2011\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=1\\x=-4021\end{array} \right.\)
Vậy `x∈{1;-4021}`
`b)2<|x-2011|<5`
`→x∈{3;4}`
`→` \(\left[ \begin{array}{l}|x-2011|=3\\|x-2011|=4\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x-2011=3\\x-2011=-3\\x-2011=4\\x-2011=-4\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=2014\\x=2008\\x=2015\\x=2007\end{array} \right.\)
Vậy `x∈{2007;2008;2014;2015}`
`2`
`a)2.2^(x+5)=128`
`→2^(x+5)=64`
`→2^(x+5)=2^6`
`→x+5=6`
`→x=1`
Vậy `x=1`
`b)3.3^x=81`
`→3^x=27`
`→3^x=3^3`
`→x=3`
Vậy `x=3`
`c)25+5^x . 5^x=650`
`→5^(2x)=625`
`→5^(2x)=5^4`
`→2x=4`
`→x=2`
Vậy `x=2`
`d)3^x . 3^(x+1) - 81=162`
`→3^(2x+1)=243`
`→3^(2x+1)=3^5`
`→2x+1=5`
`→2x=4`
`→x=2`
Vậy `x=2`
