Question

A= (\(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\)) : (\(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\))

tìm ĐKXĐ và rút gọn

gúp vs tối nạp rồi

Answer 1

ĐKXĐ: \(a\ge0,a\ne1\)

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)

\(A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)}\right)\)

\(A=\dfrac{a-1}{\sqrt{a}\left(a-1\right)}:\left(\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)}\right)\)

\(A=\dfrac{1}{\sqrt{a}}.\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}\)

\(A=\dfrac{1}{\sqrt{a}}.\left(\sqrt{a}-1\right)=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Answer 2

ĐKXĐ:\(\left\{{}\begin{matrix}\sqrt{a}>0\\\sqrt{a}-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

A= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\) ĐK:a >0;a\(\ne1\)

= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

= \(\left(\dfrac{\sqrt{a}.\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

= \(\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

= \(\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{1}{\sqrt{a}-1}\)

= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)

= \(\dfrac{a-1}{\sqrt{a}}\)

Vậy A= \(\dfrac{a-1}{\sqrt{a}}\) với a >0; a\(\ne1\)

Answer 3

A=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

ĐKXĐ a lớn hơn 0 và a khác 1

A=\(\left(\dfrac{a-1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}}\right).\left(\sqrt{a}-1\right)\)

=\(\dfrac{a-1}{\sqrt{a}}\)

* có gì không hiểu thì hỏi nha