\(\frac{4^8.3^{12}.27^2}{6^{12}.9^3}\)

= \(\frac{\left(2^2\right)^8.3^{12}.27^2}{\left(2.3\right)^{12}.\left(3^2\right)^3}\)

= \(\frac{2^{16}.3^{12}.27^2}{2^{12}.3^{12}.27^2}\)

= \(\frac{2^{16}}{2^{12}}\)= 2⁴ = 16

\(\Leftrightarrow3^x=5.3^{12}+4.\left(3^3\right)^4=5.3^{12}+4.3^{12}.\)

\(\Leftrightarrow3^x=9.3^{12}=3^2.3^{12}=3^{14}\Leftrightarrow x=14\)

bang 14

106,8 : 12 + 14,7

= 8,9 + 14,7

=  23,6

332,1 - 27 x 5,2

= 332,1 - 140,4

= 191,7

d, 468,7 : 10 x 0,1

= 46,87 x 0,1 

= 4,687

\(\frac{12}{27}+\frac{5}{8}+\frac{2}{5}+\frac{15}{27}+\frac{3}{5}+\frac{3}{8}\)

\(=\left(\frac{12}{27}+\frac{15}{27}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\)

\(=1+1+1\)

\(=3\)

(12/27+15/27)+(5/8+3/8)+3/5

=1+1+3/5

=2+3/5

=13/5

chúc em học tốt

      \(\frac{12}{27}+\frac{5}{8}+\frac{2}{5}+\frac{15}{27}+\frac{3}{5}+\frac{3}{8}\)

\(=\left(\frac{12}{27}+\frac{15}{27}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\)

\(=1+1+1\)

\(=3\)

Lời giải:
$5x+12\vdots x-2$

$\Rightarrow (5x-10)+22\vdots x-2$

$\Rightarrow 5(x-2)+22\vdots x-2$

$\Rightarrow 22\vdots x-2$

$\Rightarrow x-2\in\left\{1; -1; 2;-2;11;-11;22;-22\right\}$

$\Rightarrow x\in\left\{3; 1; 4; 0; 13; -9; 24; -20\right\}$

Đặt \(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)

\(A=x^3+27-54-x^3\)

\(A=27\)

Thay x = 27 vào biểu thức , ta có : A = 27

Vậy........................

Phá ra t thấy 

biểu thức = x^3-3x^2+9x+3x^2-9x+27-(54+x^3)

= x^3+27 -54-x^3

=-27.

Dmd , đen V . Sai dấu !

\(27^x=3^{x+2}\)

=>\(\left(3^3\right)^x=3^{x+2}\)

=>\(3^{3x}=3^{x+2}\)

=>\(3x=x+2\)

=>\(x+1\)

\(27^x=3^{x+2}\)

\(3^{3x}=3^{x+2}\)

\(\Rightarrow x+2=3x\)

\(\Rightarrow2=2x\)

\(\Rightarrow x=1\)

Vậy ...

Trả lời:

\(27^x=3^x+2\)

\(27^x-3^x=2\)

\(\left(3^3\right)^x-3^x=2\)

\(3^{3x}-3^x=2\)

\(3^{2x}=2\) (vô lí)

Vậy không có giá trị nào của x t/m yêu cầu đề bài

~ Chắc zậy á, mak cx k chắc lắm =w=

#Huyền Anh