Question

A = \(\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\) ; Rút Gọn A 
Ai chỉ mình cách làm với ạ 

Answer 1

DKXD a>0

ở tử có chứa mẫu

\(a^2+\sqrt{a}=\sqrt{a}.\left(\sqrt{a^3}+1\right)=\sqrt{a}.\left(\sqrt{a}+1\right)\left(a-\sqrt{a}11\right)\)

cái kia cũng thế

Answer 2

\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(ĐK:x>0\right)\)

\(=\frac{a^2+\sqrt{a}}{\left(a-\sqrt{a}+1\right)}-2\sqrt{a}-1+1\)

\(=\frac{a^2+\sqrt{a}-2a\sqrt{a}+2a-2\sqrt{a}}{a-\sqrt{a}+1}\)

\(=\frac{a^2-2a\sqrt{a}+2a-\sqrt{a}}{a-\sqrt{a}+1}=\frac{\sqrt{a}.\left(\sqrt{a}-1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}=\sqrt{a}\left(\sqrt{a}-1\right)\)