
Answer:
Solve for x , 1xβββ2+1xβ+β2=4(xβββ2)(xβ+β2) .
1xβββ2+1xβ+β2=4(xβββ2)(xβ+β2)
(xβββ2)(xβ+β2)(xβββ2)+(xβββ2)(xβ+β2)(xβ+β2)=4(xβββ2)(xβ+β2)(xβββ2)(xβ+β2)
(xβ2)+(x+2)=4
2x=4
x=2
But 2 is excluded from the domain of the original equation because it would make the denominator of one of the fractions zero--and division by zero is not allowed! Β . Β Therefore, it cannot be a root of the original equation. Β So, 2 is an extraneous solution. So, the equation has no solutions.
Step-by-step explanation: