a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

Sửa đề: \(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

ĐKXĐ: x<>1/2

\(\dfrac{2x-1}{3}=\dfrac{27}{2x-1}\)

=>\(\left(2x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`

`<=>x ne 4,x me -1`

`b,ĐKXĐ:4x^2-25 ne 0`

`<=>(2x-5)(2x+5) ne 0`

`<=>x ne +-5/2`

`c,ĐKXĐ:8x^3+27 ne 0`

`<=>8x^3 ne -27`

`<=>2x ne -3`

`<=>x ne -3/2`

`d,2x+2 ne 0,4y^2-9 ne 0`

`<=>2x ne -2,(2y-3)(2y+3) ne 0`

`<=>x ne -1,y ne +-3/2`

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)

bạn ghi rõ đề câu b ra đi ạ

a)\(\dfrac{x}{27}=-\dfrac{2}{3,6}\)

\(\Rightarrow\)\(3,6x=27.(-2)\)

\(\Leftrightarrow\)\(3,6x=54\)

\(x=-15\)

a, \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

=> 6(2x-7) = 9(x-3)

=> 12x - 42 = 9x - 27

=> 12x - 9x = -27 + 42

=> 3x = 15 

=> x = 5

Vậy x = 5

b, \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

=> -7(x + 27) = 6(x + 1)

=> -7x - 189 = 6x + 6

=> -7x - 6x = 6 + 189

=> -13x = 195

=> x = -15

Vậy x = -15

a) Ta có: \(\dfrac{6}{x-3}=\dfrac{9}{2x-7}\)

\(\Leftrightarrow6\left(2x-7\right)=9\left(x-3\right)\)

\(\Leftrightarrow12x-42=9x-27\)

\(\Leftrightarrow12x-9x=-27+42\)

\(\Leftrightarrow3x=15\)

hay x=5

Vậy: x=5

b) Ta có: \(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Leftrightarrow6\left(x+1\right)=-7\left(x+27\right)\)

\(\Leftrightarrow6x+6=-7x+189\)

\(\Leftrightarrow6x+7x=189-6\)

\(\Leftrightarrow13x=183\)

hay \(x=\dfrac{183}{13}\)

Vậy: \(x=\dfrac{183}{13}\)

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}+\dfrac{1}{2}\)

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)

\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

\(\dfrac{x^2-x}{x^2-1}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2}\\ \dfrac{3x}{x^3+2x^2+x}=\dfrac{3x}{x\left(x^2+2x+1\right)}=\dfrac{3}{\left(x+1\right)^2}\\ 2x=\dfrac{2x\left(x+1\right)^2}{\left(x+1\right)^2}\)

Lời giải:

a) 

$3^{2x+1}.7^y=9.21^x=3^2.(3.7)^x=3^{2+x}.7^x$

Vì $x,y$ là số tự nhiên nên suy ra $2x+1=2+x$ và $y=x$

$\Rightarrow x=y=1$

b) \(\frac{27^x}{3^{2x-y}}=\frac{3^{3x}}{3^{2x-y}}=3^{x+y}=243=3^5\Rightarrow x+y=5(1)\)

\(\frac{25^x}{5^{x+y}}=\frac{5^{2x}}{5^{x+y}}=5^{x-y}=125=5^3\Rightarrow x-y=3\) $(2)$

Từ $(1);(2)\Rightarrow x=4; y=1$