Question

cho biểu thức A=\(\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\);B=\(\dfrac{2\sqrt{a}}{\sqrt{a}+3}\)-\(\dfrac{\sqrt{a}}{3-\sqrt{a}}\)-\(\dfrac{3a+3}{a-9}\)

a) tìm ĐKXĐ của A và B

b)rút gọn B

c) đặt P=\(\dfrac{A}{B}\).Tìm a để P>\(\dfrac{1}{3}\)

d) tìm a nguyên để Q=\(\dfrac{5P\sqrt{a}}{3}\) nhận giá trị là số nguyên

Answer 1

a: ĐKXĐ: a>=0; a<>9

b: \(B=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{a-9}\)

\(=\dfrac{-3\sqrt{a}-3}{a-9}\)

Answer 2

`a)`\(ĐKXĐ:\left\{{}\begin{matrix}a\ge0\\a\ne9\end{matrix}\right.\)

`b)`\(B=\dfrac{2\sqrt{a}}{\sqrt{a}+3}-\dfrac{\sqrt{a}}{3-\sqrt{a}}-\dfrac{3a+3}{a-9}\)

\(B=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)+\sqrt{a}\left(\sqrt{a}+3\right)-3a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(B=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(B=\dfrac{-3\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)

\(B=-\dfrac{3}{\sqrt{a}-3}\)

`c)`\(P=\dfrac{A}{B}\)

\(P=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}:-\dfrac{3}{\sqrt{a}-3}\)

\(P=-\dfrac{\sqrt{a}+1}{\sqrt{a}-3}.\dfrac{\sqrt{a}-3}{3}\)

\(P=-\dfrac{\sqrt{a}+1}{3}\)

`@`\(P>\dfrac{1}{3}\)

\(\Leftrightarrow-\dfrac{\sqrt{a}+1}{3}>\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{a}+1< -1\) 

\(\Leftrightarrow\sqrt{a}< -2\) ( vô lý )

`d)`\(Q=\dfrac{5P\sqrt{a}}{3}\)

\(Q=-\dfrac{5\dfrac{\sqrt{a}+1}{3}.\sqrt{a}}{3}\)

\(Q=-\dfrac{5a+5\sqrt{a}}{9}\)

\(\Leftrightarrow5a+5\sqrt{a}+9Q=0\)

`@`TH1: Nếu \(Q=0\) \(\Rightarrow a=0\) `(tm)`

`@`TH2: Nếu \(Q\ne0\)

\(\Delta=5^2-4.5.9=-155< 0\) `=>` vô lý

Vậy \(a=0\) thì `Q` nhận giá trị nguyên