What is (sqrt(x^my^n))/(2x^3y) ?

The solution to (sqrt(x^my^n))/(2x^3y) is (x^{(-6+m)/2}y^{(-2+n)/2})/(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (sqrt(x^my^n))/(2x^3y)

simplify

\frac{x ^{m} y ^{n}}{2 x ^{3} y}

\frac{x ^{\frac{- 6 + m}{2}} y ^{\frac{- 2 + n}{2}}}{2}

Solution steps

\frac{x ^{m} y ^{n}}{2 x ^{3} y}

Apply radical rule:

ab = a b,

assuming

a \geq 0, b \geq 0

x^{m} y^{n} = x^{m} y^{n}

= \frac{x ^{m} y ^{n}}{2 x ^{3} y}

Simplify

x^{m} : x^{\frac{m}{2}}

= \frac{x ^{\frac{m}{2}} y ^{n}}{2 x ^{3} y}

Apply exponent rule:

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

\frac{x ^{\frac{m}{2}}}{x ^{3}} = x^{\frac{m}{2} - 3}

= \frac{x ^{\frac{m}{2} - 3} y ^{n}}{2 y}

Simplify

y^{n} : y^{\frac{n}{2}}

= \frac{x ^{\frac{m}{2} - 3} y ^{\frac{n}{2}}}{2 y}

Apply exponent rule:

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

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

= \frac{x ^{\frac{m}{2} - 3} y ^{\frac{n}{2} - 1}}{2}

Simplify

x^{\frac{m}{2} - 3} y^{\frac{n}{2} - 1} : x^{\frac{- 6 + m}{2}} y^{\frac{- 2 + n}{2}}

= \frac{x ^{\frac{- 6 + m}{2}} y ^{\frac{- 2 + n}{2}}}{2}

3 - 25

- \frac{1}{2} + 6

- \frac{1}{4} + 4

\frac{1}{x} + \frac{1}{x + 1}

What is (sqrt(x^my^n))/(2x^3y) ?
The solution to (sqrt(x^my^n))/(2x^3y) is (x^{(-6+m)/2}y^{(-2+n)/2})/(2)