x=7/27

\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)

\(\Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\)

\(\Leftrightarrow x=\frac{4}{9}+\frac{2}{9}\)

\(\Leftrightarrow x=\frac{2}{3}\)

Vậy\(x=\frac{2}{3}\)

Mink

\(\left(\frac{3}{4}\right)^{3x-1}=\left(\frac{16}{9}\right)^2=>3x-1=2\)2

\(\frac{5^2\cdot25^4}{125^3}=\frac{5^2\cdot\left(5^2\right)^4}{\left(5^3\right)^3}=\frac{5^2\cdot5^8}{5^9}=\frac{5^{10}}{5^9}=5\)

a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó (2a + 3c)(2b - 3d) 

= (2bk + 3dk)(2b - 3d)

= k(2b + 3d)(2b - 3d) (1)

(2a - 3c)(2b + 3d)

= (2bk - 2dk)(2b + 3d)

= k(2b - 3d)(2b + 3d) (2)

Từ (1)(2) => (2a + 3c)(2b - 3d) = (2a - 3c)(2b + 3d)

b) Sửa đề (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d) 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có (4a + 3b)(4c - 3d) = (4bk + 3b)(4dk - 3d) = bd(4k + 3)(4k - 3) (1)

Lại có (4a - 3b)(4c + 3d) = (4bk - 3b)(3dk + 3d) = bd(4k- 3)(4k + 3) (2)

Từ (1)(2) => (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d) 

1, Ta có: \(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{2a}{2b}=\frac{3c}{3d}=\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)

\(\Rightarrow\left(2a+3c\right).\left(2b-3d\right)=\left(2a-3c\right).\left(2b+3d\right)\)

        Vậy (2a + 3c).(2b - 3d) = (2a - 3c).(2b + 3d)

Câu 2 cũng tương tự nên tự làm đi