Ta có : 3^x + 3^{x + 2} = 810

=> 3^x(1 + 3²) = 810

=> 3^x.10 = 810

=> 3^x = 81

=> 3^x = 3⁴

=> x = 4

ta có \(3^3+3^x+2=810\)

=>\(3^x\left(1+3^2\right)=810\)

=>\(3^x.10=810\)

=>\(3^x=81\)

=>\(3^x=3^4\)

=>x=4

Vậy x=4

a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)

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

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

b) Đề sai ! Sửa :

\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)

\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

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

c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)

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

\(\Leftrightarrow x+3=\frac{4}{3}\)

\(\Leftrightarrow x=-\frac{5}{3}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)

\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left(3^3.\left(3-1\right)\right)^2}=\frac{3^6}{3^6.2^2}=\frac{1}{4}.\)

con thứ 2 làm tương tự. hình như là đề con thứ 2 sai em ơi

Đề 2 e chỉnh lại \(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{64}{125}\)

a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)

\(VP=18^{16}=2^{16}\cdot3^{32}\)

=> VT=VP

b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)

(đề sai)

c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)

\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)

12⁸.9¹²=18⁶

=2¹⁶.3⁸.3²⁴=2¹⁶.3³²

=2¹⁶.3³²=18⁶

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}=\left(\frac{-4}{5}\right)^3\)

=> \(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)

=> \(\frac{2}{3}x=\frac{3}{5}-\frac{-4}{5}\)

=> \(\frac{2}{3}.x=\frac{3}{5}+\frac{4}{5}=\frac{7}{5}\)

=> \(x=\frac{7}{5}:\frac{2}{3}\)

=> \(x=\frac{7}{5}.\frac{3}{2}=\frac{21}{10}\)

giúp mình với

\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}=\left(\frac{-4}{5}\right)^3\)

\(\Rightarrow\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)

\(\Rightarrow\frac{2}{3}x=\frac{3}{5}-\frac{-4}{5}=\frac{7}{5}\)

\(\Rightarrow x=\frac{7}{5}.\frac{3}{2}=\frac{21}{10}\)

Ta có:

\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{\left(5^3\right)^3.\left(5-1\right)^3}{\left(5^3\right)^5}=\frac{5^9.4^3}{5^{15}}=\frac{4^3}{5^6}=\frac{64}{5^6}\)  (1)

\(\frac{64}{25^3}=\frac{64}{\left(5^2\right)^3}=\frac{64}{5^6}\)  (2)

Từ (1) và (2) =>\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^3}\)

Bằng nhau