\[ \begin{array}{l} =2^{6} \cdot 3^{1} \\ =192 \end{array} \] \[ \begin{array}{l} =\frac{2^{x}(3)}{3.2^{x}} \\ =1 \end{array} \] EXERCISE 1 Simplify the expressions without using a calculator: (1) \( \frac{1}{4} a^{5} b^{12} \times \frac{16}{5} a^{3} b^{2} \) \( 2 \frac{2\left(2 a^{4}\right)^{3}}{16 a^{12}} \) (3) \( \frac{6 a^{4}+9 a^{4}}{5 a^{2}} \) \( 4 \quad \frac{3 a^{3} \times 5 a^{5}}{\left(5 a^{3}\right)^{2}} \) \( 5 p^{x-1} p^{x+1} \) (6) \( -2\left(-4 a^{2}+b^{3}\right)^{2} \) \( 7 \quad 16^{4 x} \div 4^{2 x} \) (8) \( \frac{12^{x+1} \cdot 27^{x-2}}{18^{2 x-1}} \) 9) \( \frac{2^{2010} \times 5^{2011}}{10^{1000}} \) \( 10 \frac{3^{x}+3^{x+1}}{8} \) Negative exponents and the meaning of \( a^{0} \) \( a^{0}=1 \) because \( 5 \div 5=1 \) and \( 5^{1} \div 5^{1}=\frac{5^{1}}{5^{1}}=5^{1-1}=5^{0}=1 \quad \) Using Law 2 For example: \( (2 x)^{0}=1 ; 2 x^{0}=2 \times 1=2 ;(a+b)^{0}=1 \)

Let's simplify each expression step by step. ### (1) \( \frac{1}{4} a^{5} b^{12} \times \frac{16}{5} a^{3} b^{2} \) First, we can multiply the coefficients and the variables separately: \[ \frac{1}{4} \times \frac{16}{5} = \frac{16}{20} = \frac{4}{5} \] Now for the variables: \[ a^{5} \times a^{3} = a^{5+3} = a^{8} \] \[ b^{12} \times b^{2} = b^{12+2} = b^{14} \] Combining these results, we have: \[ \frac{4}{5} a^{8} b^{14} \] ### (2) \( 2 \frac{2(2 a^{4})^{3}}{16 a^{12}} \) First, simplify the expression inside the parentheses: \[ (2 a^{4})^{3} = 2^{3} (a^{4})^{3} = 8 a^{12} \] Now substitute this back into the expression: \[ 2 \frac{2 \cdot 8 a^{12}}{16 a^{12}} = 2 \frac{16 a^{12}}{16 a^{12}} = 2 \cdot 1 = 2 \] ### (3) \( \frac{6 a^{4}+9 a^{4}}{5 a^{2}} \) Combine the terms in the numerator: \[ 6 a^{4} + 9 a^{4} = 15 a^{4} \] Now simplify: \[ \frac{15 a^{4}}{5 a^{2}} = 3 a^{4-2} = 3 a^{2} \] ### (4) \( \frac{3 a^{3} \times 5 a^{5}}{(5 a^{3})^{2}} \) First, simplify the denominator: \[ (5 a^{3})^{2} = 25 a^{6} \] Now simplify the numerator: \[ 3 a^{3} \times 5 a^{5} = 15 a^{3+5} = 15 a^{8} \] Now combine: \[ \frac{15 a^{8}}{25 a^{6}} = \frac{15}{25} a^{8-6} = \frac{3}{5} a^{2} \] ### (5) \( p^{x-1} p^{x+1} \) Using the property of exponents: \[ p^{(x-1) + (x+1)} = p^{2x} \] ### (6) \( -2\left(-4 a^{2}+b^{3}\right)^{2} \) First, expand the square: \[ (-4 a^{2} + b^{3})^{2} = 16 a^{4} - 8 a^{2} b^{3} + b^{6} \] Now multiply by \(-2\): \[ -2(16 a^{4} - 8 a^{2} b^{3} + b^{6}) = -32 a^{4} + 16 a^{2} b^{3} - 2b^{6} \] ### (7) \( 16^{4 x} \div 4^{2 x} \) Convert to the same base: \[ 16 = 4^{2} \Rightarrow 16^{4x} = (4^{2})^{4x} = 4^{8x} \] Now simplify: \[ \frac{4^{8x}}{4^{2x}} = 4^{8x - 2x} = 4^{6x} \] ### (8) \( \frac{12^{x+1} \cdot 27^{x-2}}{18^{2 x-1}} \) Convert to prime bases: \[ 12 = 2^{2} \cdot 3^{1}, \quad 27 = 3^{3}, \quad 18 = 2^{1} \cdot 3^{2} \] Now express each term: \[ 12^{x+1} = (2^{2} \cdot 3^{1})^{x+1} = 2^{2(x+1)} \cdot 3^{x+1} \] \[ 27^{x-2} = (3^{3})^{x-2} = 3^{3(x-2)} = 3^{3x - 6} \] \[ 18^{2x-1} = (2^{1} \cdot 3^{2})^{2x-1} = 2^{2x-1} \cdot 3^{2(2x-1)} = 2^{2x-1} \cdot 3^{4x-2} \] Now substitute back: \[ \frac{2^{2(x+1)} \cdot 3^{x+1} \cdot 3^{3x-6}}{2^{2x-1} \cdot 3^{4x-2}} = \frac{2^{2x+2} \cdot 3^{4x-5}}{2^{2x-1} \cdot 3^{4x-2}} \] Now simplify: \[ = 2^{(2x+2) - (2x-1)} \cdot 3^{(4x-5) - (4x-2)} = 2^{3} \cdot 3^{-3} = \frac{8}{27} \] ### (9) \( \frac{2^{2010} \times 5^{2011}}{10^{1000}} \) Convert \(10^{1000}\): \[ 10^{1000} = (2 \cdot 5)^{1000} = 2^{1000} \cdot 5^{1000} \] Now simplify: \[ \frac{2^{2010} \cdot 5^{2011}}{2^{1000} \cdot 5^{1000}} = 2^{2010-1000} \cdot 5^{2011-1000} = 2^{1010} \cdot 5^{1011} \] ### (10) \( \frac{3^{x}+3^{x+1}}{8} \) Factor out \(3^{x}\): \[ = \frac{3^{x}(1 + 3)}{8} = \frac{3^{x} \cdot 4}{8} = \frac{3^{x}}{2} \] ### Summary of Results 1. \( \frac{4}{5} a^{8}