Answer:
Assume x greater-than-or- equal-to 0 and y ... 0 StartRoot x squared y cubed EndRoot + 2 StartRoot x cubed y Superscript 4 Baseline EndRoot + x y StartRoot y EndRoot.
Step-by-step explanation:
Option (b) is correct.
The sum of given expression \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy is 2xy\sqrt{y}+2xy^2\sqrt{x}2xyy+2xy2x
Step-by-step explanation:
Given:Expression \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy
We have to find the sum of the given expression and choose the correct from the given options.
Consider the given expression \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy .
\sqrt{x^2y^3}x2y3 can be written as \sqrt{x^2y^2y}=xy\sqrt{y}x2y2y=xyy
Also, 2\sqrt{x^3y^4}2x3y4 can be written as 2\sqrt{x^3y^4}=2\sqrt{x^2x(y^2)^2}=2xy^2\sqrt{x}2x3y4=2x2x(y2)2=2xy2x
Now, the given expression becomes,
x2y3+2x3y4+xyy
=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}=xyy+2xy2x+xyy .
Now, adding like term, terms having same variable with same degree.
=2xy\sqrt{y}+2xy^2\sqrt{x}=2xyy+2xy2x .
Thus, The sum of given expression \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy is 2xy\sqrt{y}+2xy^2\sqrt{x}2xyy+2xy2x
