Question

giải chi tiết câu b cho giúp mình với ạ

 

Answer 1

a:

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

 \(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\cdot\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(a-1\right)}{a-1}\cdot\dfrac{a-1}{\sqrt{a}}\)

\(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{\sqrt{a}}\)

\(=\dfrac{4\sqrt{a}+4\sqrt{a}\left(a-1\right)}{\sqrt{a}}\)

=4+4(a-1)

=4a

b: \(a=\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)\sqrt{2-\sqrt{3}}\)

\(=\left(2\sqrt{3}-2+3-\sqrt{3}\right)\cdot\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\left(\sqrt{3}+1\right)\cdot\dfrac{\left(\sqrt{3}-1\right)}{\sqrt{2}}=\dfrac{3-1}{\sqrt{2}}=\sqrt{2}\)

Khi \(a=\sqrt{2}\) thì \(P=4\cdot\sqrt{2}=4\sqrt{2}\)