Question

Solve the following:
$\frac{x-1}{x^{3/4}+x^{1/2}}\cdot \frac{x^{1/2}+x^{1/4}}{x^{1/2}+1}\cdot x^{1/4}+1$ for $x=16$

Answer 1

Here the value of $x=16$ so , put the value of $x$ in the given equation we

$\frac{16-1}{{16}^{\frac{3}{4}}+{16}^{\frac{1}{2}}}$.$\frac{{16}^{\frac{1}{2}}+{16}^{\frac{1}{4}}}{{16}^{\frac{1}{2}}+1}$.${16}^{\frac{1}{4}}+1$

$\frac{16-1}{(2^4{)}^{\frac{3}{4}}+(2^4{)}^{\frac{1}{2}}}$.$\frac{(2^4{)}^{\frac{1}{2}}+(2^4{)}^{\frac{1}{4}}}{(2^4{)}^{\frac{1}{2}}+1}$.$(2^4{)}^{\frac{1}{4}}+1$

$\frac{16-1}{2^3+2^2}$.$\frac{2^2+2}{2^2+1}$.$2+1$

$\frac{15}{12}$. $\frac{6}{5}.2+1$

$\therefore$ The answer is $3+1=4$