1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)

\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)

\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)

\(\Rightarrow3^x=3\)

\(\Rightarrow x=1\)

2) \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)

\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=-46\)

3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)

\(\Rightarrow2^{2x-3}=2^{49}\)

\(\Rightarrow2x-3=49\)

\(\Rightarrow2x=52\)

\(\Rightarrow x=26\)

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

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

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

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

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

________

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

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

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

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

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

_____________

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

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

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

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

\(x=2\)

________

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

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           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=\(\dfrac{15}{4}\)}

3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)

\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)

Vậy x = -7 hoặc x = 11

4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)

\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)

5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)

\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)

a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)

b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)

c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)

\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

Lời giải:

Gọi biểu thức là $A$

\(A=\frac{2^{10}.3^8+5.(2^2)^5.3^8}{2^{10}.(3^3)^3-2^{10}.(3^2)^4}=\frac{2^{10}.3^8+5.2^{10}.3^8}{2^{10}.3^9-2^{10}.3^{8}}\)

\(=\frac{2^{10}.3^8(1+5)}{2^{10}.3^8(3-1)}=\frac{6}{2}=3\)

(3x+21)³=\(\dfrac{27}{8}\)

\(\Rightarrow\)(3x+21)³=\(\left(\dfrac{3}{2}\right)^3\)

\(\Rightarrow\)3x+21=\(\dfrac{3}{2}\)

\(\Rightarrow\)3x =\(\dfrac{3}{2}\)-21=\(\dfrac{-39}{2}\)

\(\Rightarrow\)x =\(\dfrac{-39}{2}\):3

\(\Rightarrow\) =\(\dfrac{-13}{2}\)

Vậy x =\(\dfrac{-13}{2}\)

\(\left(3x+21\right)^3=\dfrac{27}{8}\)

\(\Rightarrow3x+3.7=\dfrac{1}{2}.3\)

\(\Rightarrow x+7=\dfrac{1}{2}\)

\(\Rightarrow x=-7\dfrac{1}{2}\)

Giải:

\(\left(3x+21\right)^3=\dfrac{27}{8}\)

\(\Leftrightarrow\left(3x+21\right)^3=\left(\dfrac{3}{2}\right)^3\)

Vì \(3=3\)

Nên \(3x+21=\dfrac{3}{2}\)

\(\Leftrightarrow3x=\dfrac{3}{2}-21\)

\(\Leftrightarrow3x=-\dfrac{39}{2}\)

\(\Leftrightarrow x=-\dfrac{39}{2}:3\)

\(\Leftrightarrow x=-\dfrac{13}{2}\)

Vậy \(x=-\dfrac{13}{2}\).

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

a. Kiểm tra lại mẫu số vế phải, \(7-5x\) hay \(7-3x\) 

b. ĐKXĐ: \(x\ne-\dfrac{5}{3}\)

\(\dfrac{3x+5}{12}=\dfrac{3}{5+3x}\)

\(\Leftrightarrow\dfrac{\left(3x+5\right)^2}{12\left(3x+5\right)}=\dfrac{36}{12\left(3x+5\right)}\)

\(\Rightarrow\left(3x+5\right)^2=36=6^2\)

\(\Rightarrow\left[{}\begin{matrix}3x+5=6\\3x+5=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\) (thỏa mãn)