a,(5x-1)⁶=3⁶

5x-1=3

x=4/5

b,(2x+1)³=0,1³

2x+1=0,1

x=-0,45

c,(2x-3)⁴=(2x-3)⁴(2x-3)²

(2x-3)²=0

2x-3=0

x=3/2

d,(2x+1)⁵=(2x+1)⁵(2x+1)²⁰⁰⁵

(2x+1)²⁰⁰⁵=0

2x+1=0

x=-1/2

a)\(\orbr{\begin{cases}5x-1=3\\5x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=4\\5x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{-2}{5}\end{cases}}\)

b) 2x+1=-0,1 <=> 2x=-1,1=>x=-0,55

c) (2x-3)⁴ .[1-(2x-3)² ]=0

do (2x-3)⁴ lớn hơn 0 nên 1-(2x-3)²=0=>(2x-3)²=1=>2x-3=1=>2x=4=>x=2

d) tương tự câu c)

a) 5x-1=3 hoặc 5x-1=-3 sau đo tính ra

bổ sung câu c) chỗ cuối đấy (2x-3)²=1=>2x-3=1 hoặc 2x-3=-1 sau tính đc x=1 hoặc x=2

(2x+1)³=729
=>2x+1=\(\sqrt[3]{729}\)
=>2x+1=9
=>2x=8
=>x=4

tick đúng nha

3^x+3*3^2x-1*3^x=729

\(\Rightarrow3^{x+3+2x-1+x}=3^6\)

\(\Rightarrow3^{4x+2}=3^6\)

\(\Rightarrow4x+2=6\)

\(\Rightarrow4x=4\)

\(\Rightarrow x=1\)

Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)

\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)

\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)

Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)

\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)

\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)

Vì \(\frac{-3}{2}< 0\)nên A < B

\(3^{2x}=\left(3^2\right)^x\)\(\Rightarrow9^x=729\)\(\Rightarrow x=3\)

\(3^{2x}\) = 729

<=>\(9^x\) = 729

<=>9 . 9 .9 = 729

<=>  \(9^3\) = 729

<=>   x        =3

Vay :x=3

3^2x = 729 

9^x=729

 Ta có : 9.9.9=729

=> x = 3

\(3^{2x-1}=\frac{1}{729}=\frac{1}{3^6}=3^{-6}\Rightarrow2x-1=-6\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)

\(\text{Vậy:}x=-\frac{5}{2}\)