How to solve $(1 - a^{2}) \div ((\frac{1 - a^{\frac{3}{2}}}{1 - a^{\frac{1}{2}}} + a^{\frac{1}{2}}) \cdot (\frac{1 + a^{\frac{3}{2}}}{1 + a^{\frac{1}{2}}} - a^{\frac{1}{2}})) + 1$ $(1-a^2)÷(((1-a^(3/2))/(1-a^(1/2))+a^(1/2))*((1+a^(3/2))/(1+a^(1/2))-a^(1/2)))+1$ , $a \geq 0, a \neq 1$ $a>=0, a≠1$ ?

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Answer 1 · 2018-05-05T19:00:44

$\frac{2}{1 - a}, (a \geq 0, a \neq 1)$ $2/(1-a), (a ge0, a!=1)$ .

Let, $a^(1/2)=x. :. a^(3/2)=x^3$ .

$:. (1-a^(3/2))/(1-a^(1/2))$ ,

$=(1-x^3)/(1-x)$ ,

$={cancel((1-x))(1+x+x^2)}/cancel((1-x))$ .

$:. (1-a^(3/2))/(1-a^(1/2))=(1+x+x^2)$ .

$:. (1-a^(3/2))/(1-a^(1/2))+a^(1/2)=(1+x+x^2)+x=1+2x+x^2, or,$

$(1-a^(3/2))/(1-a^(1/2))+a^(1/2)=(1+x)^2.......(star^1)$ .

On the similar lines, we can have,

$(1+a^(3/2))/(1+a^(1/2))-a^(1/2)=(1-x)^2........(star^2)$ .

Combining $(star^1) and (star^2)$ , we have,

${((1-a^(3/2))/(1-a^(1/2))+a^(1/2))( (1+a^(3/2))/(1+a^(1/2))-a^(1/2))}=(1-x^2)^2$ .

Hence, the Exp. $=(1-x^4)-:(1-x^2)^2+1$ ,

$={(1+x^2)(1-x^2)}-:(1-x^2)^2+1$ ,

$=(1+x^2)/(1-x^2)+1$ ,

$={(1+x^2)+(1-x^2)}/(1-x^2)$ ,

$=2/(1-x^2)$ ,

$=2/(1-a), (a ge 0, a!=1)$ .

How to solve (1-a^2)÷(((1-a^(3/2))/(1-a^(1/2))+a^(1/2))*((1+a^(3/2))/(1+a^(1/2))-a^(1/2)))+1(1−a2)÷((1−a321−a12+a12)⋅(1+a321+a12−a12))+1(1-a^2)÷(((1-a^(3/2))/(1-a^(1/2))+a^(1/2))*((1+a^(3/2))/(1+a^(1/2))-a^(1/2)))+1, a>=0, a≠1a≥0,a≠1a>=0, a≠1 ?