Bạn coi kĩ lại câu 1 đi bạn \(\sqrt{x}-2\) chứ không phải \(\sqrt{x-2}\)
Câu 1:
Với \(x>0,x\ne4\), ta có:
\(A=\left(\dfrac{x+2\sqrt{x}}{x-2\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{2}{\sqrt{x}-2}\)
b) Với \(x>0,x\ne4\): \(A< 0\)
\(\Rightarrow\dfrac{2}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\left(2>0\right)\)
\(\Leftrightarrow x< 4\)
Câu 2:
\(A=\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\)
\(\Rightarrow\sqrt{2}A=\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\)
\(\Rightarrow\sqrt{2}A=\sqrt{17}+1-\sqrt{17}+1\)
\(\Rightarrow\sqrt{2}A=2\)
\(\Rightarrow A=\sqrt{2}\)
câu 2 : Ta có
\(\sqrt{2}A=\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{17+2\sqrt{17}+1}-\sqrt{18-2\sqrt{17}+1}\)\(\Leftrightarrow\sqrt{2}A=\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}=\left|\sqrt{17}+1\right|-\left|\sqrt{17}-1\right|\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{17}+1-\sqrt{17}+1=2\)
\(\Leftrightarrow A=\sqrt{2}\)
