Question

\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{11\sqrt{x}-3}{9-x}\) \(\left(x\ge0,x\ne9\right)\)

a, Rút gọn A

b, Tìm A khi x =25

c,Tìm x để giá trị tuyệt đối của A lớn hơn A

Answer 1

\(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(x=25\Rightarrow A=\frac{3.5}{5-3}=\frac{15}{2}\)

Để \(\left|A\right|>A\Rightarrow A< 0\Rightarrow\sqrt{x}-3< 0\Rightarrow x< 9\Rightarrow0\le x< 9\)