3^x+1 + 2x.3^x - 18x - 27 = 0

=> 3^x.3 + 2x.3^x - (18x + 27) = 0

=> 3^x.(3 + 2x) - 9.(2x + 3) = 0

=> (2x + 3).(3^x - 9) = 0

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\3^x-9=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=-3\\3^x=9\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=2\end{cases}}\)

Vậy ...

<=> 3^x(2x+3)-9(2x+3)=0

<=> (2x+3)(3^x-9)=0

<=> 2x+3=0 => x=-3/2

Và: 3^x-9=0 => 3^x=9=3² => x=2

Đs: x=-3/2 và x=2

\(3^{x+1}+2x.3^x-18x-27=0\)

\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3^x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=2\end{matrix}\right.\)

Vậy ...............

3^x.3+2x.3^x-(18x+27)=0

=> 3^x(3+2x)-9.(3+2x)=0

=> (3^x-9).(3+2x)=0

=> \(\left[{}\begin{matrix}3^x-9=0\\3+2x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3^x=9=3^2\\2x=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(3^{x+1}+2x+3^x-18x-27=0\)

<=> \(3^x\left(3+2x\right)-9\left(2x+3\right)=0\)

<=> \(\left(2x+3\right)\left(3^x-9\right)=0\)

<=>\(\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

vậy.......

3^{x + 1} + 2x .3^x - 18x - 27 = 0

<=> 3^x ( 3 + 2x ) - 9 ( 2x + 3 ) = 0

<=> ( 3^x - 9 ) ( 2x + 3 ) = 0

<=> \(\orbr{\begin{cases}3^x-9=0\\2x+3=0\end{cases}}\)<=>\(\orbr{\begin{cases}3^x=9\\2x=-3\end{cases}}\)

<=>\(\orbr{\begin{cases}3^x=3^2\\x=-\frac{3}{2}\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)

3^x+1+2X.3^x-18x+27=0

<=> 3^x(3+2x)-9(3+2x)=0

<=> (3+2x)(3^x-9)=0

\(\Leftrightarrow\orbr{\begin{cases}3+2x=0\\3^x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1,5\\x=2\end{cases}}\)

Cảm ơn

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)