\(=\dfrac{3^5}{-\left(0.003\right)^3\cdot0.09}=-10^{11}\)

mình cũng dag bí

\(\frac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\frac{\left(0,125.2,4\right)^5}{-1.\left(0,3\right)^5.\left(0,01\right)^3}=\frac{0,3^5}{\left(0,3\right)^5.\left(-0,01\right)^3}=\frac{1}{\left(-0,01\right)^3}\)

Có \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)=\(\left(\dfrac{-5}{4}+\dfrac{2}{4}\right)^2\)=\(\left(\dfrac{-3}{4}\right)^2\)=\(\dfrac{\left(-3\right)^2}{4^2}=\dfrac{9}{16}\)

Có \(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,125.2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,3\right)^5}{\left(-0.3\right)^5.\left(0,01\right)^3}=\dfrac{1}{-1.\left(0,01\right)^3}=\dfrac{1}{-\left(0,01\right)^3}\)

a/ \(\frac{25^2\times4^2}{5^5\times\left(-2\right)^5}\)= \(\frac{625\times16}{3125\times\left(-32\right)}\)=\(\frac{10000}{-10000}\)= -1

a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=2^5\)

b) \(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{5^4.2^4}{5^5.\left(-2\right)^5}=\dfrac{1.1}{5.\left(-2\right)}=\dfrac{1}{-10}=\dfrac{-1}{10}\)

( Mình giải được có 2 bài thôi. ) :)

a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=3^5\)

b)\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(25.4\right)^2}{\left(5.\left(-2\right)\right)^5}=\dfrac{\left(100\right)^2}{\left(-10\right)^5}=\dfrac{1}{-10}\)

Còn phần c mình cx đang tắc

\(=\dfrac{\left(0.125\cdot2.4\right)^5}{\left(-0.3\right)^3\cdot\left(0.01\right)^3\cdot\left(-0.3\right)^2}=-\dfrac{\left(0.3\right)^5}{\left(0.3\right)^5\cdot\left(0.01\right)^3}\)

\(=-\dfrac{1}{\left(0.01\right)^3}=-\dfrac{1}{\dfrac{1}{1000000}}=-1000000\)