\( 27 \left(\dfrac{1}{9}\right)^2 \left(\dfrac{9^2}{3^5} \right) = \)
Solution
Use the quotient rule of exponents \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite \( \left(\dfrac{1}{9}\right)^2 \) in the given expression as
\( 27 \left(\dfrac{1}{9}\right)^2 \left(\dfrac{9^2}{3^5} \right) = 27 \; \dfrac{1^2}{9^2} \; \dfrac{9^2}{3^5} \)
Rewrite \( 27 \) as \( 3^3 \)
\( 3^3 \; \dfrac{1^2}{9^2} \; \dfrac{9^2}{3^5} \)
Simplify \( 1^2 = 1 \), remove brackets, multiply and rewrite as one fraction,
\( = \dfrac{3^3 9^2}{9^2 3^5} \)
Divide the numerator and the denominator by the GCF \(= 3^3 9^2\) of the two and simplify (or cancel common factors)
\( = \dfrac{1}{3^2} = \dfrac{1}{9} \)
\( 80 \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 = \)
Solution
Use rules of exponents \( a^m \cdot b^m = (a \cdot b)^m \)
to rewrite \( \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 \) as \( \left(\dfrac{1}{5^{-1}} \dfrac{25^{-1}} {4} \right)^2 \) in the given the expression
\( 80 \left(\dfrac{1}{5^{-1}}\right)^2 \left(\dfrac{25^{-1}}{4}\right)^2 = 80 \left(\dfrac{1}{5^{-1}} \dfrac{25^{-1}} {4} \right)^2 \)
Change negative exponents to positive exponents using the rule \( a^{-n} = \dfrac{1}{a^n} \) to rewrite \( 5^{-1} = \dfrac {1}{5} \) and \( 25^{-1} = \dfrac{1}{25} \)
\( = 80 \left(\dfrac{1}{\dfrac{1}{5}} \dfrac{\dfrac{1}{25}} {4} \right)^2 \)
Use division of fractions
to rewrite \( \dfrac{1}{\dfrac{1}{5}} = 5 \) and \( \dfrac{\dfrac{1}{25}} {4} = \dfrac{1}{25 \times 4} = \dfrac{1}{100} \) to rewrite the expression as
\( = 80 \left( 5 \times \dfrac{1}{100} \right)^2\)
Multiply inside brackets
\( = 80 \left( \dfrac{5}{100} \right)^2\)
Divide numerator and denominator by \( 5 \) and simplify
\( = 80 \left( \dfrac{1}{20} \right)^2\)
Use the rule of exponents \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite \( \left( \dfrac{1}{20} \right)^2\) in the given expression
\( = 80 \times \dfrac{1^2}{20^2} \)
Simplify
\( = \dfrac{80}{400} \)
Divide numerator and denominator by the common factor \( 80 \) to simplify the given expression to
\( = \dfrac {1}{5} \)
For x and y not equal to zero, the expression
\( \left(\dfrac{x^4}{y^5}\right)^3 \left(\dfrac{y^2}{x^2}\right)^2 \)
simplifies to
Solution
Use the rule of exponents \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) and remove brackets to rewrite the given expression as
\( \left(\dfrac{x^4}{y^5}\right)^3 \left(\dfrac{y^2}{x^2}\right)^2 = \dfrac{(x^4)^3}{(y^5)^3} \dfrac{(y^2)^2}{(x^2)^2} \)
Use the rule of exponents \( (a^n)^m = a^{n \cdot m} \) to rewrite the expression as
\( = \dfrac{x^{12}}{y^{15}} \dfrac{y^4}{x^4} \)
Apply the quotient rule \( \dfrac{a^m}{a^n} = a^{m - n} \) to simplify
\( = \dfrac{x^{12-4}}{ y^{15-4} } = \dfrac{x^8}{ y^{11}}\)
For \( x \ne - y \),
\( \dfrac{3(2x + 2y)^5}{4 (x + y)^3} = \)
Solution
Factor \( 2 \) out in the expression \( 2x + 2y \) and write given expression as
\( \dfrac{3(2x + 2y)^5}{4 (x + y)^3} = \dfrac{3(2(x + 2)y)^5}{4 (x + y)^3} \)
Apply exponent rule \( (a \cdot b)^m = a^m \cdot b^m \) to write expression as
\( = \dfrac{3 \times 2^5(x + 2)^5}{4 (x + y)^3} \)
Rearrange as
\( = \dfrac{3 \times 2^5}{2^4} \dfrac{(x + y)^5}{(x + y)^3} \)
Use the quotient rule of exponents \( \dfrac{a^m}{a^n} = a^{m - n} \) to rewrite the above as
\( = 3 \times 2^{5-4} \times (x + y)^{5-3} \)
Simplify
\( = 24 (x + y)^2 \)
For x and y not equal to zero, the expression
\( \left(\dfrac{x^0}{2y}\right)^2 \left(\dfrac{y^4}{x^3}\right)^2 \)
simplifies to
Solution
Use rules of exponents \( x^0 = 1 \) and \( a^m \cdot b^m = (a \cdot b)^m \) to rewrite given expression as
\( \left(\dfrac{x^0}{2y}\right)^2 \left(\dfrac{y^4}{x^3}\right)^2 = \left(\dfrac{1 \times y^4}{2y \times x^3}\right)^2 \)
Simplify inside brackets
\( = \left(\dfrac{y^{4-1}}{2 x^3}\right)^2 = \left(\dfrac{y^3}{2 x^3}\right)^2\)
Apply \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite the above as
\( = \dfrac{y^6}{4 x^6} \)
\( (- 3 x^2 y^3) (- 4 x^3 y^5) = \)
Solution
Multiply and put terms with the same variable within brackets
\( (- 3 x^2 y^3) (- 4 x^3 y^5) = (-3 \times (-4)) (x^2 \times x^3)(y^3 \times y^5) \)
Apply the exponent rule \( a^m \cdot a^n = a^{m+n} \)
\( = 12 (x^{2+3})(y^{3+5}) \)
Simplify
\( = 12 x^5 y^8 \)
For x and y not equal to zero, the expression
\( \dfrac{12 x^3 y^{-2}}{4 x^{-2}y^3} \)
simplifies to
Solution
Rearrange and write as the product of rational expressions with the same variable
\( \dfrac{12 x^3 y^{-2}}{4 x^{-2}y^3} = \dfrac{12}{4} \times \dfrac{x^3}{x^{-2}} \times \dfrac{y^{-2}}{y^3} \)
Apply the exponent rule \( \dfrac{a^m}{a^n} = a^{m - n} \)
\( = 3 x^{3-(-2)} )(y^{-2-3}) \)
Simplify
\( = 3 x^5 y^{-5} \)
Write with positive integer
\( = 3 \dfrac{x^5}{ y^5} \)
For x and y not equal to zero,
\( \left(\dfrac{3 x^{-2}}{y^2}\right)^{-2} = \)
For x and y not equal to zero,
Solution
Apply the exponent rule \( \left( \dfrac{a}{b} \right)^{-m} = \dfrac{b^m}{a^m} \) to rewrite the given expression as
\( \left(\dfrac{3 x^{-2}}{y^2}\right)^{-2} = \dfrac{({y^2})^2}{(3 x^{-2})^2} \)
Apply the rule exponents \( (a^n)^m = a^{n \cdot m} \) in the numerator and the exponent rule \( (a \cdot b)^m = a^m \cdot b^m \) in the denominator
\( = \dfrac{y^4}{3^2 (x^{-2})^2} \)
Simplify
\( = \dfrac{y^4}{ 9 x^{-4}} \)
Write with positive exponents
\( = \dfrac{x^4 y^4}{ 9} \)
For x and y not equal to zero, the expression
\( \left( \dfrac{2x^2 y^{-1}}{5} \right)^2 \left(\dfrac{5 x^{-1}y^3}{4}\right)^3 \)
simplifies to
Solution
Apply the exponent rule \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to rewrite the given expression as
\( \left( \dfrac{2x^2 y^{-1}}{5} \right)^2 \left(\dfrac{5 x^{-1}y^3}{4}\right)^3 = \dfrac{(2x^2 y^{-1})^2}{5^2} \dfrac{(5 x^{-1}y^3)^3}{4^3} \)
Apply the exponent rule \( (a \cdot b)^m = a^m \cdot b^m \) in the numerators
\( = \dfrac{2^2 (x^2)^2 (y^{-1})^2 }{5^2} \dfrac{5^3 (x^{-1})^3 (y^3)^3}{4^3} \)
Apply the rule exponents \( (a^n)^m = a^{n \cdot m} \) in the numerator to rewrite the above as
\( = \dfrac{2^2 x^4 y^{-2} }{5^2} \dfrac{5^3 x^{-3} y^9}{4^3} \)
Write with positive exponents using the rule of negative exponents \( a^{-n} = \dfrac{1}{a^n} \)
\( = \dfrac{2^2 x^4 }{5^2} \dfrac{1}{y^2} \dfrac{5^3 y^9}{4^3} \dfrac{1}{x^3} \)
Multiply
\( = \dfrac{2^2 x^4 }{5^2 y^2} \dfrac{5^3 y^9}{4^3 x^3} \)
Rewrite as a multiplication of rational (fractional) expressions of the same variable
\( = \dfrac{2^2 x^4 }{5^2 y^2} \dfrac{5^3 y^9}{4^3 x^3} = \dfrac{4 \times 5^3}{4^3 \times 5^2} \dfrac{x^4}{x^3} \dfrac{y^9}{y^2} \)
Apply the rule of exponents \( \dfrac{a^m}{a^n} = a^{m - n} \)
\( \dfrac{5^{3-2}}{4^{3-1}} x^{4-3} y^{9-2} \)
Simplify
\( \dfrac{5}{16} x y^7 \)
For x and y not equal to zero,
\( \left(\dfrac{x^{-1}}{y^0} \right)^2 \left(\dfrac{x^2}{y^3}\right)^3 = \)
Solution
Apply the rule of exponents \( y^0 = 1 \) and the rule \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) and simplify
\( \left(\dfrac{x^{-1}}{y^0} \right)^2 \left(\dfrac{x^2}{y^3}\right)^3 = (x^{-1})^2 \dfrac{(x^2)^3}{(y^3)^3} \)
Apply the rule exponents \( (a^n)^m = a^{n \cdot m} \) to rewrite the above as
\( = x^{-2} \dfrac{x^6}{y^9} \)
Multiply and simplify
\( = \dfrac{x^{6-2}}{y^9} \)
\( = \dfrac{x^4}{y^9} \)
For \( y \ne 0 \), the expression
\( \left(\dfrac{x^2}{y^3}\right) \left(\dfrac{8x}{y}\right) \left(\dfrac{x^2}{4}\right)^2 \)
simplifies to
Solution
Apply the rule \( \left( \dfrac{a}{b} \right)^m = \dfrac{a^m}{b^m} \) to the term on the right
\( \left(\dfrac{x^2}{y^3}\right) \left(\dfrac{8x}{y}\right) \left(\dfrac{x^2}{4}\right)^2 = \left(\dfrac{x^2}{y^3}\right) \left(\dfrac{8x}{y}\right) \left(\dfrac{x^4}{4^2}\right) \)
Multiply
\( = \dfrac{x^2 \times 8x \times x^4}{y^3 \times y \times 16} \)
Apply the rule \( a^m \cdot a^n = a^{m+n} \)
\( = \dfrac{8 x^{2+1+4}}{16 y^{3+1}} \)
Simplify
\( = \dfrac{1}{2} \dfrac{x^7}{y^4} \)
