Question

What is the following sum? Assume x > 0 and y > 0 sqrt x^2y^2+2 sqrt x^3y^4+xy sqrt y

Answer 1

Answer:

$xy(1+2y\sqrt{x}+\sqrt{y})$

Step-by-step explanation:

Given expression,

$\sqrt{x^2y^2}+2\sqrt{x^3y^4}+xy\sqrt{y}$

$=(x^2y^2)^\frac{1}{2} + 2(x^3y^4)^\frac{1}{2} + xy\sqrt{y}$

$\because (\sqrt{x}=x^\frac{1}{2})$

$=(x^2)^\frac{1}{2} (y^2)^\frac{1}{2} + 2(x^3)^\frac{1}{2} (y^4)^\frac{1}{2} + xy\sqrt{y}$

$(\because (ab)^n=a^n b^n)$

$=x^{2\times \frac{1}{2}} y^{2\times \frac{1}{2}} + 2(x^{3\times \frac{1}{2}})(y^{4\times \frac{1}{2}})+xy\sqrt{y}$

$\because (a^n)^m=a^{mn}$

$=x^1 y^1 + 2x^{1\frac{1}{2}} y^2 + xy\sqrt{y}$

$=xy+2x.(x)^\frac{1}{2} y^2 + xy\sqrt{y}$

$=xy+2xy^2\sqrt{x}+xy\sqrt{y}$

$=xy(1+2y\sqrt{x}+\sqrt{y})$

Answer 2

Answer:

c

Step-by-step explanation:

just did it