Question

Cho biểu thức: \(A=\left(1-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)Với:x\ge0,x\ne4,x\ne9\)

a) Rút gọn A

b) Tìm các giá trị nguyên của x để A nhận giá trị nguyên

LÀM ƠN GIÚP MK VS!!!

Answer 1

b/A=\(\frac{x-2\sqrt{x}-3-3\sqrt{x}+9}{x-2\sqrt{x}-3}=1-\frac{3\left(\sqrt{x}-3\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=1-\frac{3}{1+\sqrt{x}}\)

Vậy 1+ căn x thuốc Ư(3), mà \(\sqrt{x}\ge0\Rightarrow1+\sqrt{x}\ge1\)

Vậy \(1+\sqrt{x}=\left(1,3\right)\)

\(\Rightarrow\sqrt{x}=\left(0,2\right)\) Vì x nguyên nên x=0

Answer 2

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

\(\Leftrightarrow\frac{1}{1+\sqrt{x}}:\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{1+\sqrt{x}}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{x-9-x+4+\sqrt{x}+2}\)

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

\(\Leftrightarrow A=\frac{x-5\sqrt{x}+6}{x-2\sqrt{x}-3}\)