Question

[Lê Ng Hải Anh]

Trục căn thức ở mẫu:

\(\frac{\sqrt{a+b}-\sqrt{a}+\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}\)

Answer 1

[alibaba nguyễn]

\(A=\frac{\sqrt{a+b}-\sqrt{a}+\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}=1+\frac{2\sqrt{b}}{\sqrt{a+b}-\sqrt{a}-\sqrt{b}}=1+B\)

\(B=\frac{2\sqrt{b}\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{-2\sqrt{ab}}=-\frac{\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}}=-\frac{\sqrt{a}\left(\sqrt{a+b}+\sqrt{a}+\sqrt{b}\right)}{a}\)

Answer 2

[Lê Ng Hải Anh]

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

\(=\frac{2a+2b-2\sqrt{a\left(a+b\right)}+2\sqrt{b\left(a+b\right)}-2\sqrt{ab}}{2a-2\sqrt{a\left(a+b\right)}}\)

\(=\frac{a+b-\sqrt{a\left(a+b\right)}+\sqrt{b\left(a+b\right)}-\sqrt{ab}}{a-\sqrt{a\left(a+b\right)}}\)

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

\(=\frac{b\sqrt{a\left(a+b\right)}+\sqrt{ab}\left(a+b\right)-a\sqrt{ab}}{-ab}\)

\(=\frac{-b\sqrt{a\left(a+b\right)}+b\sqrt{ab}}{ab}\)

\(=\frac{\sqrt{ab}-\sqrt{a\left(a+b\right)}}{a}\)