Question

hộ mình câu c) với ạaa

Answer 1

1: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)

2: \(A=\dfrac{1}{3}\)

=>\(\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}\)

=>\(\sqrt{x}+3=3\cdot3=9\)

=>\(\sqrt{x}=9-3=6\)

=>x=36(nhận)

3: \(\sqrt{x}+3>=3\forall x\) thỏa mãn ĐKXĐ

=>\(A=\dfrac{3}{\sqrt{x}+3}< =\dfrac{3}{3}=1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0