Lời giải:
a. $2^{x+1}.3^y=12^x=(2^2.3)^x=2^{2x}.3^x$

$\Rightarrow x+1=2x; y=x$

$\Rightarrow x=1; y=x\Rightarrow x=y=1$

b. 

$2^x=4^{y-1}$

$2^x=(2^2)^{y-1}=2^{2(y-1)}$
$\Rightarrow x=2(y-1)(1)$
Lại có:

$27^y=3^{x+8}$

$(3^3)^y=3^{x+8}$

$3^{3y}=3^{x+8}\Rightarrow 3y=x+8(2)$

Từ $(1); (2)\Rightarrow 3y=2(y-1)+8$

$\Leftrightarrow 3y=2y+6$

$\Leftrightarrow y=6$

$x=2(y-1)=2(6-1)=10$

Đề đã đầy đủ chưa bạn? Và bạn đang cần làm gì với biểu thức này? 

tih bieu thuc ma

\((2x-3)^3=(2x-3)^5\\\Rightarrow (2x-3)^3-(2x-3)^5=0\\\Rightarrow (2x-3)^3[1-(2x-3)^2]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-3\right)^3=0\\1-\left(2x-3\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x-3=1\\2x-3=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\2x=4\\2x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

\(Vậy:x\in\left\{1;\dfrac{3}{2};2\right\}\)

`#3107.101107`

`|3x - 1| = x + 2`

`\Rightarrow` TH1: `3x - 1 = x + 2`

`\Rightarrow 3x - x = 2 + 1`

`\Rightarrow 2x = 3`

`\Rightarrow x =` $\dfrac{3}2$

TH2: `3x - 1 = -(x + 2)`

`\Rightarrow 3x - 1 = -x - 2`

`\Rightarrow 3x + x = -2 + 1`

`\Rightarrow 4x = -1`

`\Rightarrow x =` $\dfrac{-1}4$

Vậy, \(x\in\left\{-\dfrac{1}{4};\dfrac{3}{2}\right\}.\)

|3x - 1| = x + 2

*) TH1: x ≥ 1/3, ta có:

|3x - 1| = x + 2

3x + 1 = x + 2

3x - x = 2 - 1

2x = 1

x = 1/2 (nhận)

*) TH2: x < 1/3, ta có:

|3x - 1| = x + 1

1 - 3x = x + 1

-3x - x = 1 - 1

-4x = 0

x = 0 (nhận)

Vậy x = 0; x = 1/2