Question

How do I solve this, pretty urgent please. ​

Answer 1

Answer:

• 1

Step-by-step explanation:

Substitute:

• $x^{0.5} = y$

Then the expression becomes:

• $(\dfrac{1+y^3}{1-y+y^2}-y) \dfrac{1-y^2}{1-y} -y=$
• $(\dfrac{(1+y)(1-y+y^2)}{1-y+y^2}-y) \dfrac{(1-y)(1+y)}{1-y} -y=$
• $(1+y-y)(1+y)-y=$
• $1(1+y)-y=$
• $1+y-y=$
• 1

The result is 1 regardless the value of x other than 1.