Question

A.\(\sqrt{4x+8}-\sqrt{9x+18}+\sqrt{x+2}=\sqrt{x+5}\) Giai phương trình

b, rút gọn biểu thức: (\(\dfrac{1}{1+2\sqrt{x}}-\dfrac{1}{\sqrt{3}+2}\))/\(\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)

Answer 1

a) \(\sqrt{4x+8}-\sqrt{9x+18}+\sqrt{x+2}=\sqrt{x+5}\)

\(\Leftrightarrow\sqrt{4\left(x+2\right)}-\sqrt{9\left(x+2\right)}+\sqrt{x+2}=\sqrt{x+5}\)

\(\Leftrightarrow2\sqrt{x+2}-3\sqrt{x+2}+\sqrt{x+2}=\sqrt{x+5}\)

\(\Leftrightarrow0\sqrt{x+2}=\sqrt{x+5}\Leftrightarrow0=\sqrt{x+5}\)

\(\Leftrightarrow0=x+5\Leftrightarrow-5=x\)

Vậy phương trình đã cho có nghiệm duy nhất là x = -5

b) ĐKXĐ: \(x\ge0;x\ne1\)

\(T=\left(\dfrac{1}{1+2\sqrt{x}}-\dfrac{1}{\sqrt{3}+2}\right):\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)

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

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

Answer 2

a) Bổ sung: ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x+2}XĐ\Leftrightarrow x+2\ge0\\\sqrt{x+5}XĐ\Leftrightarrow x+5\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\ge-5\end{matrix}\right.\Rightarrow}x\ge-2}\) Sau khi tìm được x = -5 ta thấy k thỏa mãn Đk: \(x\ge-2\)

Vậy pt đã cho là vô nghiệm