Bài 2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{x+\sqrt{x}}{\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}-x+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
|x+1|=5
=>\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-6\left(loại\right)\end{matrix}\right.\)
Thay x=4 vào A, ta được:
\(A=\dfrac{2\cdot\sqrt{4}}{\sqrt{4}-1}=\dfrac{2\cdot2}{2-1}=4\)
