Question

[Kimi No Nawa]

Rút gọn biểu thức:

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\left(a,b>0\right)\)

Giúp mk nha =)))) Đừng vt đáp án là đc

Answer 1

[💋Bevis💋]

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

\(=\frac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

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

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

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

\(=2\sqrt{b}\)

Answer 2

[KAl(SO4)2·12H2O]

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{-b+\sqrt{a}.\sqrt{b}}{\sqrt{b}}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right].\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right).\sqrt{b}}-\frac{\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right]-\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\sqrt{a}+2b.\sqrt{b}}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=\frac{2b.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}\)

\(D=2\sqrt{b}\)