Question

Tìm ĐK và rút gọn

a,\(A=\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)

b, \(B=\left(\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\right):\dfrac{1}{x^2-\sqrt{x}}\)

Answer 1

a) điều kiện xác định : \(-1< x< 1\)

ta có : \(A=\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)

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

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

b) điều kiện xác định : \(x>0;x\ne1\)

ta có : \(B=\left(\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\right):\dfrac{1}{x^2-\sqrt{x}}\)

\(\Leftrightarrow B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\right):\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) \(\Leftrightarrow B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\right).\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

\(\Leftrightarrow B=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)=x-1\)

Answer 2

Phùng Khánh Linh