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Đáp án:

Giải thích các bước giải:

 Bài `1`:
$\left(-\frac{1}{4}\right)^0 = 1$
 $\left(-2\frac{1}{3}\right)^2 = \left(-\frac{7}{3}\right)^2 = \frac{(-7)^2}{3^2} = \frac{49}{9}$
$(0,5)^3 = \left(\frac{1}{2}\right)^3 = \frac{1^3}{2^3} = \frac{1}{8}$
$\left(-1\frac{1}{3}\right)^4 = \left(-\frac{1 \cdot 3 + 1}{3}\right)^4 = \left(-\frac{4}{3}\right)^4 = \frac{(-4)^4}{3^4} = \frac{256}{81}$

Bài `2`: 
a) $\left(\frac{4}{5}\right)^5 \cdot x = \left(\frac{4}{5}\right)^7$
$x = \left(\frac{4}{5}\right)^7 \div \left(\frac{4}{5}\right)^5$
$x = \left(\frac{4}{5}\right)^{7-5}$
$x = \left(\frac{4}{5}\right)^2$
$x = \frac{4^2}{5^2}$
$x = \frac{16}{25}$

b) $(3x+1)^3 = -27$
$(3x+1)^3 = (-3)^3$
$3x+1 = -3$
$3x = -3-1$
$3x = -4$
$x = -\frac{4}{3}$

Bài `3:`

a) $(-5)^{30}$ và $(-3)^{50}$
Ta có:
$(-5)^{30} = ((-5)^3)^{10} = (-125)^{10} = 125^{10}$
$(-3)^{50} = ((-3)^5)^{10} = (-243)^{10} = 243^{10}$
Vì $125 < 243$, nên $125^{10} < 243^{10}$.
Vậy, $(-5)^{30} < (-3)^{50}$.

b) $64^8$ và $16^{12}$
Ta có:
$64^8 = (2^6)^8 = 2^{6 \cdot 8} = 2^{48}$
$16^{12} = (2^4)^{12} = 2^{4 \cdot 12} = 2^{48}$
Vì $2^{48} = 2^{48}$.
Vậy, $64^8 = 16^{12}$.

Bài 4: 

a) $27^3 \cdot 3^2$
$27^3 \cdot 3^2 = (3^3)^3 \cdot 3^2 = 3^{3 \cdot 3} \cdot 3^2 = 3^9 \cdot 3^2 = 3^{9+2} = 3^{11}$
$3^{11} = 177147$

b) $\left(\frac{3}{5}\right)^{15} \div \left(\frac{9}{25}\right)^5$
$\left(\frac{3}{5}\right)^{15} \div \left(\frac{3^2}{5^2}\right)^5$
$= \left(\frac{3}{5}\right)^{15} \div \left(\left(\frac{3}{5}\right)^2\right)^5$
$= \left(\frac{3}{5}\right)^{15} \div \left(\frac{3}{5}\right)^{10}$
$= \left(\frac{3}{5}\right)^{15-10}$
$= \left(\frac{3}{5}\right)^5$
$= \frac{3^5}{5^5} = \frac{243}{3125}$

c) $5 - \left(-\frac{5}{11}\right)^0 + \left(\frac{1}{3}\right)^2 : 3$
$= 5 - 1 + \frac{1^2}{3^2} : 3$
$= 5 - 1 + \frac{1}{9} : 3$
$= 4 + \frac{1}{27}$
$= \frac{108}{27} + \frac{1}{27} = \frac{109}{27}$

d) $2^3 + 3 \cdot \left(\frac{1}{2}\right)^0 + [(-2)^2 : \frac{1}{2}] \cdot 8$
$= 8 + 3 \cdot 1 + [4 : \frac{1}{2}] \cdot 8$
$= 8 + 3 + [8] \cdot 8$
$= 11 + 64$
$= 75$