a: \(\left(\sqrt{3}\right)^x=243\)

=>\(3^{\dfrac{1}{2}\cdot x}=3^5\)

=>\(\dfrac{1}{2}\cdot x=5\)

=>x=10

b: \(0,1^x=1000\)

=>\(\left(\dfrac{1}{10}\right)^x=1000\)

=>\(10^{-x}=10^3\)

=>-x=3

=>x=-3

c: \(\left(0,2\right)^{x+3}< \dfrac{1}{5}\)

=>\(\left(0,2\right)^{x+3}< 0,2\)

=>x+3>1

=>x>-2

d: \(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{5}{3}\right)^2\)

=>\(\left(\dfrac{3}{5}\right)^{2x+1}>\left(\dfrac{3}{5}\right)^{-2}\)

=>2x+1<-2

=>2x<-3

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

e: \(5^{x-1}+5^{x+2}=3\)

=>\(5^x\cdot\dfrac{1}{5}+5^x\cdot25=3\)

=>\(5^x=\dfrac{3}{25,2}=\dfrac{1}{8,4}=\dfrac{10}{84}=\dfrac{5}{42}\)

=>\(x=log_5\left(\dfrac{5}{42}\right)=1-log_542\)

a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

\((x-1)^5=-243\)

\(\Rightarrow x-1=(-3)^5\)

\(\Rightarrow x-1=-3\)

\(\Rightarrow x=-3+1\)

\(\Rightarrow x=-2\)

\(\left(x-1\right)^5=-243\)

\(\Rightarrow\left(x-1\right)^5=\left(-3\right)^5\)

\(\Rightarrow x-1=-3\)

\(\Rightarrow x=-3+1\)

\(\Rightarrow x=-2\)

\(\left(x-1\right)^5=-243\)
\(\left(x-1\right)^5=-3^5\)

\(x-1=-3\)

\(x=-3+1\)

\(x=-2\)

x = từ 1 đến 10000....0

\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)

=>\(\left(\frac{3}{5}\right)^x.\left(\frac{5}{3}\right)^{12}=\left(\frac{3}{5}\right)^5\)

=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{5}{3}\right)^{12}\)

=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^{17}\)

=>x=17

\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)

\(\Rightarrow\left(\frac{3}{5}\right)^x.\left(\frac{3}{5}\right)^{12}=\left(\frac{3}{5}\right)^5\)

\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{3}{5}\right)^{12}\)

\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{5}{3}\right)^7\)

\(\Rightarrow x=-7\)

a.

\(\left(x-\frac{1}{5}\right)^5=\frac{1}{243}\)

\(x-\frac{1}{5}=\sqrt[5]{\frac{1}{243}}\)

\(x-\frac{1}{5}=\frac{1}{3}\)

\(x=\frac{1}{3}+\frac{1}{5}\)

\(x=\frac{8}{15}\)

b.

|2x-1|-x=1

\(\Leftrightarrow\orbr{\begin{cases}2x-1-x=1\\-2x+1-x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

Vậy x= 0 hoặc x=2

c. \(\left|\frac{3}{5}-\frac{1}{2}x\right|>\frac{2}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{5}-\frac{1}{2}x>\frac{2}{5}\\-\frac{3}{5}+\frac{1}{2}x>\frac{2}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x< \frac{1}{5}\\\frac{1}{2}x>\frac{-1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x< \frac{2}{5}\\x>\frac{-2}{5}\end{cases}}\)

Vậy....

                                                         Bài giải

a, \(\left(x-\frac{1}{5}\right)^5=\frac{1}{243}\)

\(\left(x-\frac{1}{5}\right)^5=\left(\frac{1}{2}\right)^5\)

\(x-\frac{1}{5}=\frac{1}{2}\)

\(x=\frac{1}{2}+\frac{1}{5}\)

\(x=\frac{7}{10}\)

a)(2*x-7)^5=243

=>(2*x-7)^5=3^5

=>2*x-7=3

=>2*x=3+7

=>2*x=10

=>x=10:2

=>x=5

Vậy x=5

\(( 2.x -7 )^5 = 3^5 \)