Đáp án+Giải thích các bước giải:
`a)2^x+2^{x+1}+2^{x+2}=28`
`=>2^{x}.(1+2^1+2^2)=28`
`=>2^{x}.(1+2+4)=28`
`=>2^{x}.7=28`
`=>2^{x}=28:7`
`=>2^{x}=4`
`=>2^{x}=2^2`
`=>x=2.`
Vậy `x=2.`
`b)(2x+3)^8=(2x+3)^6`
`=>(2x+3)^8-(2x+3)^6=0`
`=>(2x+3)^6[(2x+3)^2-1]=0`
`=>[((2x+3)^6=0),((2x+3)^2-1=0):}`
`=>[(2x+3=0),((2x+3)^2=1):}`
\(⇒\left[ \begin{array}{l}2x+3=0\\2x+3=1\\2x+3=-1\end{array} \right.\)
\(⇒\left[ \begin{array}{l}2x=-3\\2x=-2\\2x=-4\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=-1\\x=-2\end{array} \right.\)
Vậy `x=-3/2` hoặc `x=-1` hoặc `x=-2.`
