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

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

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

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

=>x=17

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

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

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

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

\(\Rightarrow x=-7\)

\(4^{x+2}=16=4^2\)

=> x+ 2 = 2

X = 0

\(5^{2x+3}=3125=5^5\)

=> 2x+3 = 5

=> x =1

Mấy phần còn alji tương tự

\(4^{x+2}=16=4^2\)

\(=>x+2=2\) \(=>x=0\)
\(5^{2x+3}=3125=5^5\)

\(=>2x+3=5\) \(=>x=1\)

\(6^{x+2}=216=6^3\) 

\(=>x+2=3\) \(=>x=1\)

\(5^{2\left(x+1\right)}=625=5^4\)

\(=>2\left(x+1\right)=4\) \(=>x=1\)

\(3^{4-2x}=3^0\)

\(=>4-2x=0\) \(=>x=2\)

a. x=5/6

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

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

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

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

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

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

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

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

Bài 1:

\((1-2x)^2=9=3^2=(-3)^2\)

\(\Rightarrow \left[\begin{matrix} 1-2x=3\\ 1-2x=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=2\end{matrix}\right.\)

Bài 2:

\((x+5)^3=-64=(-4)^3\)

\(\Rightarrow x+5=-4\Rightarrow x=-9\)

Bài 3:

\((3x-5)^2=16=4^2=(-4)^2\)

\(\Rightarrow \left[\begin{matrix} 3x-5=4\\ 3x-5=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=\frac{1}{3}\end{matrix}\right.\)

Bài 4:

\((x-1)^3=27=3^3\)

\(\Rightarrow x-1=3\Rightarrow x=4\)

Bài 5:

\(x^2+x=0\Leftrightarrow x(x+1)=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.\)

Bài 6:

\(5^{x+2}=625=5^4\)

\(\Rightarrow x+2=4\Rightarrow x=2\)

Bài 7:

\(5-|3x-1|=3\)

\(\Rightarrow |3x-1|=5-3=2\)

\(\Rightarrow \left[\begin{matrix} 3x-1=2\\ 3x-1=-2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{-1}{3}\end{matrix}\right.\)

Bài 8:

\(|2x-1|+5=4\Rightarrow |2x-1|=4-5=-1\)

Điều này vô lý vì trị tuyệt đối của một số thì luôn không âm

Vậy không tồn tại $x$ thỏa mãn

Bài 9:

\(|\frac{1}{3}-x|-1=\frac{2}{3}\Rightarrow |\frac{1}{3}-x|=\frac{5}{3}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{3}-x=\frac{5}{3}\\ \frac{1}{3}-x=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-4}{3}\\ x=2\end{matrix}\right.\)

a: \(\Leftrightarrow2x-3=x\)

=>x=3

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)

=>2^x=1/8

=>x=-3

c: =>2x+7=-4

=>2x=-11

=>x=-11/2

d: =>(4x-3)^2*(4x-4)(4x-2)=0

hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)

Bài làm:

a) Ta có: \(5^{x+2}=625\)

\(\Leftrightarrow5^{x+2}=5^4\)

\(\Rightarrow x+2=4\)

\(\Rightarrow x=2\)

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

\(\Leftrightarrow\left(x-1\right)^{x+2}=\left(-1\right)^{x+2}.\left(-1\right)^2\)

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

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

\(\Rightarrow x=0\)

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

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

5^x+2=625

5^x+2=5^4

x+2=4

x=4-2

x=2

(x-1)^x+2=[(-1)^2]^x+2

(x-1)=(-1)^2

(x-1)=1

x=1+1

x=2

vậy x=2

(2x-1)^3=-8

(2x-1)^3=(-2)^3

2x-1=-2

2x=-2+1

2x=-1

x=-1:2

x=-0,5

vậy x=-0,5

vậy x=2