Question

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

Rút gọn 

Tính gtri của P khi a=2\(\sqrt{3}\) và b=\(\sqrt{3}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\end{matrix}\right.\)

Ta có: \(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

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

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

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

\(=a-b\)

Thay a = 2√3 và b = √3 vào P, ta được:

P = 2√3 - √3 = √3

Vậy...

 

 

Answer 2

a) Ta có: \(P=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\cdot\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)

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

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

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

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

\(=a-b\)

b) Thay \(a=2\sqrt{3}\) và \(b=\sqrt{3}\) vào biểu thức P=a-b, ta được:

\(P=2\sqrt{3}-\sqrt{3}=\sqrt{3}\)

Vậy: Khi \(a=2\sqrt{3}\) và \(b=\sqrt{3}\) thì \(P=\sqrt{3}\)