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

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

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

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

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

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

a) = \(\frac{7}{2}\)

b) = \(\frac{643}{64}\)

c) = 0

a) 

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

\(x=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)   

b) 

\(-\frac{1}{27}\cdot x=\frac{1}{81}\)  

\(x=\frac{1}{81}:\left(-\frac{1}{27}\right)\) 

\(x=-\frac{1}{3}\)   

c) 

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

\(x-\frac{1}{2}=\frac{1}{3}\) 

\(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)   

d) 

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

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

\(\orbr{\begin{cases}x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\\x=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\end{cases}}\)

Sửa lại câu a : ( nhìn sai số ) 

\(\frac{3^5}{5^5}\cdot x=\frac{3^7}{7^7}\)  

\(x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\) 

\(x=\frac{3^7}{7^7}\cdot\frac{5^5}{3^5}\)  

\(x=\frac{5^5\cdot3^2}{7^7}\)   

\(x=\frac{28125}{823453}\)

\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)

\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)

\(=\frac{5}{2}-\frac{8}{27}-8\)

\(=\frac{-313}{54}\)

-313/54

-9/25 nghĩa là (âm chín phần hai mươi năm ) nha bạn 

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

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

\(\Rightarrow x=4\)

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

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

\(\Leftrightarrow x=2\)

k mình nha bạn

Mk làm sai rồi. Hi

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?)

16/2^x = 2

=> 2^x . 2 = 16

     2^x      = 8

     2^3     = 2^3

=> x = 3

\(\frac{16}{2^x}=2\Rightarrow16:2^x=2\)

\(\Rightarrow2^x=16:2=8\Rightarrow2^x=2^3\Rightarrow x=3\)

\(\frac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x:81=-27\)

\(\Rightarrow\left(-3\right)^x=-27\cdot81=-2187\Rightarrow\left(-3\right)^x=\left(-3\right)^7\Rightarrow x=7\)

mk biết mỗi thế thôi. mong bạn thông cảm.

~CÁC BẠN GIÚP MK LÊN 200 ĐIỂM NHA. BẠN NÀO GIÚP THÌ MK K LẠI NHÉ.~

~THANKS~

a,\(\frac{16}{2^x}=2\)

\(\frac{2^4}{2^x}\)\(=2\)

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

\(4-x=1\)

x=3

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

\(\left(-3\right)^x=\left(-3\right)^3\)

x-4=3

x=7

c,\(8^x:2^x=4\)

 \(4^x=4\)

x=1