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Simplify the following expression completely, where [tex]\( x \geq 0 \)[/tex]:

[tex]\[ x \sqrt{5 x y^4}+\sqrt{405 x^3 y^4}-\sqrt{80 x^3 y^4} \][/tex]

Sagot :

To simplify the given expression completely:

[tex]\[ x \sqrt{5 x y^4} + \sqrt{405 x^3 y^4} - \sqrt{80 x^3 y^4} \][/tex]

we proceed through a series of steps to simplify each term and combine like terms.

1. Simplify each square root term:

[tex]\[ x \sqrt{5 x y^4} = x \cdot \sqrt{5} \cdot \sqrt{x} \cdot \sqrt{y^4} = x \cdot \sqrt{5} \cdot \sqrt{x} \cdot y^2 = x \cdot y^2 \sqrt{5x} \][/tex]

[tex]\[ \sqrt{405 x^3 y^4} = \sqrt{405} \cdot \sqrt{x^3} \cdot \sqrt{y^4} = \sqrt{405} \cdot x^{3/2} \cdot y^2 = \sqrt{405} \cdot (x \cdot \sqrt{x}) \cdot y^2 = y^2 \sqrt{405 x^3} \][/tex]

Since:

[tex]\[ \sqrt{405 x^3} = \sqrt{405} \cdot x \cdot \sqrt{x} \][/tex]

So:

[tex]\[ \sqrt{405 x^3 y^4} = y^2 \cdot \sqrt{405 x^3} = y^2 \cdot \sqrt{405 x} = y^2 \sqrt{405 x} \sqrt{x} \][/tex]

[tex]\[ = \sqrt{405} x^{3/2} y^2= y^2 x \sqrt{405x} \][/tex]

[tex]\[ = x y^2\sqrt{405x} \][/tex]

[tex]\[ 405 = 5^4 \cdot 2 \cdot 81= (5\cdot 81.01 + 4 x 5) \][/tex]

[tex]\(\sqrt{80 x^3 y^4} = y^2) if we normalize then we get 80 x implicitl 2. Combine and simplify terms: Let's add these expressions together: \[ x y^2 \sqrt{5 x} + x y^2 \sqrt{405 x} x y^2(-\sqrt{80 x}) \] Observing that each term includes a common factor of \(y^2\sqrt{x}\)[/tex]:

\[
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