Question

câu 1

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

với a>0, a khác 4

a) rút gon biểu thức a

b) tìm a để A = \(\sqrt{a}+2\)

Answer 1

a) \(A=\left(\dfrac{\sqrt{a}+2}{2-\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}+2}-\dfrac{4a+2\sqrt{a}-4}{4-a}\right):\left(\dfrac{-2}{2-\sqrt{a}}+\dfrac{2+\sqrt{a}}{2\sqrt{a}-a}\right)=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\left[\dfrac{-2\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}+\dfrac{2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\right]=\left[\dfrac{a+4\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}+\dfrac{2\sqrt{a}-a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}-\dfrac{4a+2\sqrt{a}-4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}\right]:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{a+4\sqrt{a}+4+2\sqrt{a}-a-4a-2\sqrt{a}+4}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{2-\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}=\dfrac{-4a+4\sqrt{a}+8}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\dfrac{1}{\sqrt{a}}=\dfrac{4\left(2-\sqrt{a}\right)\left(\sqrt{a}+1\right).\sqrt{a}}{\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)}=\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}\)

Ta có A=\(\sqrt{a}+2\Leftrightarrow\dfrac{4a+4\sqrt{a}}{\sqrt{a}+2}=\sqrt{a}+2\Leftrightarrow4a+4\sqrt{a}=\left(\sqrt{a}+2\right)^2\Leftrightarrow4a+4\sqrt{a}=a+4\sqrt{a}+4\Leftrightarrow3a=4\Leftrightarrow a=\dfrac{4}{3}\left(tm\right)\)Vậy a=\(\dfrac{4}{3}\) thì A=\(\sqrt{a}+2\)

Answer 2

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

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

\(=\dfrac{-4a+8\sqrt{a}}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}\) .\(\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)

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

=\(\dfrac{4a}{2+\sqrt{a}}\)

b, Để A=\(\sqrt{a}+2\)

<=> \(\dfrac{4a}{2+\sqrt{a}}\) =\(\sqrt{a}+2\)

<=> 4a=\(\left(\sqrt{a}+2\right)^2\)

<=> \(a+4\sqrt{a}+4-4a=0\)

<=> \(-3a+4\sqrt{a}+4=0\)

<=>\(-3a+6\sqrt{a}-2\sqrt{a}+4=0\)

<=> \(-3\sqrt{a}\left(\sqrt{a}-2\right)-2\left(\sqrt{a}-2\right)=0\)

<=> \(\left(\sqrt{a}-2\right)\left(-3\sqrt{a}-2\right)=0\)

<=>\(\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{-2}{3}\left(vl\right)\end{matrix}\right.\)

<=> a=4