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

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

\(x=\left(\frac{3}{7}\right)^2\)

\(x=\frac{9}{49}\)

Vậy...

b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)

\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)

\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)

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

Vậy...

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

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

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

\(x=\frac{5}{6}\)

Vậy...

d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)

=>\(x+\frac{1}{4}=\frac{2}{3}\)

\(x=\frac{2}{3}-\frac{1}{4}\)

\(x=\frac{5}{12}\)

Vậy...

Phù, mãi mới xong, tk cho mk nha bn

a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)

phân tích kết quả ra bạn nhé

Dễ thấy (\(\frac{3}{4}\)-81); (\(\frac{3^2}{5}\)-81); (\(\frac{3^3}{6}\)-81);... (\(\frac{3^{2007}}{2010}\)-81) có dạng (\(\frac{3^x}{3+x}\)-81) và x\(\varepsilon\){1;2;3;...2007}.

Nếu x=6 thì \(\frac{3^x}{3+x}\)-81=\(\frac{3^6}{3+6}\)-81=0

=>  (\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...​(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)=0

Mà |x-30|-6001=(\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...​(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)

=>|x-30|-6001=0

=>|x-30|=6001

=>x-30=6001 hoặc x-30=-6001

=>x=6031 hoặc x=-5971

-------------------The end----------------

\(\text{|x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\text{ |x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^6}{9}-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\left|x-30\right|- 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(3^4-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow|x - 30| - 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...0...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\text{|x - 30| - 6001 = }0\)

\(\Rightarrow\left|x-30\right|=6001\) 

\(\Rightarrow x-30=6001\)hoặc \(x-30=-6001\)

\(\Rightarrow x=6031\)hoặc\(x=-5971\)

Vậy: x= 6031 hoặc x= -5971

(Nói thật thì mình mới lớp 7, đây có phải của lớp 8 không?)

\(\left(\frac{2}{3}\right)^{x-2}=\left(\frac{16}{81}\right)^{x+1}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{x-2}=\left[\left(\frac{2}{3}\right)^4\right]^{x+1}\)

\(\Leftrightarrow\left(\frac{2}{3}\right)^{x-2}=\left(\frac{2}{3}\right)^{4\left(x+1\right)}\)

\(\Leftrightarrow x-2=4x+4\)

\(\Leftrightarrow-3x=6\Leftrightarrow x=-2\)

=>x - 3 0 = 6001 hoặc x - 30 = -6001

=> x = 6031 hoặc x = -5971

BÀi nay không khó lắm 

 Dễ thấy vế bên phải bằng 0 vì \(\frac{3^6}{9}-81=0\)

=> lx - 30 l - 6001 = 0 

=> lx - 30 l = 6001 

Tự làm tiếp 

\(\left(-\frac{2}{3}\right)^{x-1}=\frac{16}{81}\)

\(\left(-\frac{2}{3}\right)^{x-1}=\frac{\left(-2\right)^4}{3^4}\)

\(\left(-\frac{2}{3}\right)^{x-1}=\left(-\frac{2}{3}\right)^4\)

x - 1 = 4

x = 4 + 1

x = 5

64=4^3

ta có 4^3/4^x+1=4

suy ra x=1

x = từ 1 đến 10000....0