\(\dfrac{81}{125}=\dfrac{3^4}{5^3}\) ; \(-\dfrac{8}{27}=-\left(\dfrac{2}{3}\right)^3\)

a: \(\dfrac{-11}{81}=\dfrac{1}{27}\cdot\dfrac{-11}{3}\)

b: \(\dfrac{-11}{81}=\dfrac{1}{27}:\dfrac{3}{-11}\)

a.\(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

c,\(\frac{25}{49}=\left(\frac{5}{7}\right)^2\)

\(-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)

\(\dfrac{81}{625}=\left(\dfrac{3}{5}\right)^4\)

\(\dfrac{25}{49}=\left(\dfrac{5}{7}\right)^2\)

b: \(3^4\cdot3^5:\dfrac{1}{27}==3^9\cdot3^3=3^{12}\)

\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)

\(\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

\(\dfrac{-125}{64}=\left(-\dfrac{5}{4}\right)^3\)

\(0=0^2\)

\(-27=\left(-3\right)^3\)

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

a: Sửa đề: 3^2

\(=3^2\cdot\dfrac{1}{3^5}\cdot3^8\cdot\dfrac{1}{3^3}=3^2\)

b: \(=3^{\left(-2\right)\cdot\left(-2\right)}\cdot\dfrac{1}{3^5}\cdot3^3=\dfrac{3^4}{3^2}=3^2\)

c: \(=2^{12}\cdot2^{16}\cdot2^4=2^{32}\)

d: \(=\left[\dfrac{1}{9}\cdot\dfrac{27}{8}\cdot3\right]\cdot\dfrac{128}{81}\)

\(=\dfrac{16}{9}=\left(\dfrac{4}{3}\right)^2\)

Bài 1 : bạn cứ đóng ngoặc bài lại rồi cho thêm mũ nào đó vào là xong

bài 2: 

a,(2x-3)^2=4

(2x-3)^2=(+-2)^2

=> 2x-3=(+-2)

(bn cứ phân ra 2 trường hợp rồi từ từ làm