\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(\Rightarrow x-1=4\)

\(x=4+1\)

\(x=5\)

Vậy \(x=5\)

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

\(\left(-3\right)^x=-27.81\)

\(\left(-3\right)^x=-2187\)

\(\left(-3\right)^x=\left(-3\right)^7\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

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

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

\(\Rightarrow x=4\)

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

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

\(\Leftrightarrow x=2\)

k mình nha bạn

Mk làm sai rồi. Hi

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

b) (x – 4). (x2 + 4x + 16) – x. (x2 - 6) = 2

⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – (x3 - 6x ) – 2= 0

⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – x3 + 6x – 2= 0

⇔ 6x – 66 =0

⇔ 6x = 66

⇔ x = 66 : 6

⇔ x = 11

Vậy x = 11

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

\(\left(x+\frac{1}{2}\right)^4=\frac{16}{81}\)

\(\Rightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\\\left(x+\frac{1}{2}\right)^4=\left(\frac{-2}{3}\right)^4\end{cases}}\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{3}\\x+\frac{1}{2}=\frac{-2}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-7}{6}\end{cases}}\)

Vậy  \(x\in\left\{\frac{1}{6};\frac{-7}{6}\right\}\)

_Chúc bạn học tốt_

a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.

b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .

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?)

Cách viết khác : \(\frac{16}{81}=\frac{2^4}{3^4}=\left(\frac{2}{3}\right)^4\)

Chúc bạn học tốt ^^

\(\left(\frac{16}{81}\right)=\left(\frac{2}{3}\right)^4\)

\(\frac{16}{81}=\left(\frac{-4}{-9}\right)^2=\left(\frac{2^2}{3^2}\right)^2=\left(\frac{\left(-2\right)^4}{\left(-3\right)^4}\right)=...\)