3*8^x +4*12^x -18^x -2*27^x=0

Đáp án:

$S =\{1\}$

Giải thích các bước giải:

$\quad 3.8^x + 4.12^x - 18^x - 2.27^x = 0$

$\Leftrightarrow 3.\left(\dfrac23\right)^{3x} + 4.\left(\dfrac23\right)^{2x}- \left(\dfrac23\right)^{x}- 2 = 0$

$\Leftrightarrow \left[\begin{array}{l}\left(\dfrac23\right)^x = -1\quad (loại)\\\left(\dfrac23\right)^x = \dfrac23\quad (nhận)\end{array}\right.$

$\Leftrightarrow x = 1$

Vậy $S =\{1\}$

`3.8^x+4.12^x-18^x-2. 27^x=0`

`<=> 3.(2^3)^x+4.(2^2. 3)^x-(2.3^2)^x-2.(3^3)^x=0`

`<=> 3. 2^(3x)+4. 2^(2x). 3^x-2^x. 3^(2x)-2. 3^(3x)=0`

Đặt `2^x=a; 3^x=b` ta có pt ẩn phụ a;b là:

`3a^3+4a^2 b-ab^2-2b^3=0`

`<=> 3a^3-2a^2 b+6a^2 b-4ab^2+3ab^2-2b^3=0`

`<=> a^2(3a-2b)+2ab(3a-2b)+b^2(3a-2b)=0`

`<=> (3a-2b)(a^2+2ab+b^2)=0`

`<=> (3a-2b)(a+b)^2=0`

`<=>`\(\left[ \begin{array}{l}3a-2b=0\\a+b=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}a=\dfrac{2}{3}. b(TM)\\a=-b(l)\end{array} \right.\) 

`-> 2^x=2/3. 3^x`

`<=> (2/3)^x=2/3`

`<=> x=1` 

Vậy `S={1}`