Question

Please simplify as much as possible

Answer 1

-13vx
`√(2x)`

Explanation:

v
`\sqrt{50^(3) }` - 6x
`\sqrt{18xv^(2) }`

Rewrite 50x³ as (5x)²⋅ (2x)

v
`\sqrt{(5x)^(2)(2x) }` - 6x
`\sqrt{18xv^(2) }`

Pull terms out from under the radical.

v (5x
`√(2x)`) - 6x
`\sqrt{18xv^(2) }`

Rewrite using the commutative property of multiplication.

5vx
`√(2x)` - 6x
`\sqrt{18xv^(2) }`

Rewrite 18xv² as (3v)² · (2x)

5vx(
`√(2x)`) - 18vx (
`√(2x)`)

-13vx
`√(2x)`

So, the answer is -13vx
`√(2x)`