What is [(9x^{1/2}y^{-3})(2x^{-1}y)^2]^{1/2} ?

The solution to [(9x^{1/2}y^{-3})(2x^{-1}y)^2]^{1/2} is 6/(x^{3/4)y^{1/2}}

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simplify [(9x^{1/2}y^{-3})(2x^{-1}y)^2]^{1/2}

simplify

[(9 x^{\frac{1}{2}} y^{- 3}) (2 x^{- 1} y)^{2}]^{\frac{1}{2}}

\frac{6}{x ^{\frac{3}{4}} y ^{\frac{1}{2}}}

Solution steps

((9 x^{\frac{1}{2}} y^{- 3}) (2 x^{- 1} y)^{2})^{\frac{1}{2}}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

((9 x^{\frac{1}{2}} y^{- 3}) (2 x^{- 1} y)^{2})^{\frac{1}{2}} = (9 x^{\frac{1}{2}} y^{- 3})^{\frac{1}{2}} ((2 x^{- 1} y)^{2})^{\frac{1}{2}}

= (9 x^{\frac{1}{2}} y^{- 3})^{\frac{1}{2}} ((2 x^{- 1} y)^{2})^{\frac{1}{2}}

Simplify

((2 x^{- 1} y)^{2})^{\frac{1}{2}} : \frac{2 y}{x}

= (9 x^{\frac{1}{2}} y^{- 3})^{\frac{1}{2}} \frac{2 y}{x}

(9 x^{\frac{1}{2}} y^{- 3})^{\frac{1}{2}} = \frac{3 x ^{\frac{1}{4}}}{y ^{\frac{3}{2}}}

= \frac{2 y}{x} \cdot \frac{3 x ^{\frac{1}{4}}}{y ^{\frac{3}{2}}}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{3 x ^{\frac{1}{4}} \cdot 2 y}{y ^{\frac{3}{2}} x}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{6 y x ^{\frac{1}{4}}}{x y ^{\frac{3}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{x ^{\frac{1}{4}}}{x} = \frac{1}{x ^{1 - \frac{1}{4}}}

= \frac{6 y}{y ^{\frac{3}{2}} x ^{- \frac{1}{4} + 1}}

Subtract the numbers:

1 - \frac{1}{4} = \frac{3}{4}

= \frac{6 y}{x ^{\frac{3}{4}} y ^{\frac{3}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{y}{y ^{\frac{3}{2}}} = \frac{1}{y ^{\frac{3}{2} - 1}}

= \frac{6}{x ^{\frac{3}{4}} y ^{\frac{3}{2} - 1}}

Subtract the numbers:

\frac{3}{2} - 1 = \frac{1}{2}

= \frac{6}{x ^{\frac{3}{4}} y ^{\frac{1}{2}}}

4 x^{2} + 4 - 5 x + x - 2 x^{2} + 8

23 \cdot 35

(x - 1) (x - 9)

(a + b + c + d) (a + b - c - d)

\frac{x ^{2} - 1}{x ^{2} - 2 x + 1} - 1

What is [(9x^{1/2}y^{-3})(2x^{-1}y)^2]^{1/2} ?
The solution to [(9x^{1/2}y^{-3})(2x^{-1}y)^2]^{1/2} is 6/(x^{3/4)y^{1/2}}