+) ta có : \(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
+) ta có : \(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1=1\)
+) điều kiện : \(x\ge1\)
ta có : \(C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}C=2\sqrt{x-1}\left(x\ge2\right)\\C=2\left(1\le x< 2\right)\end{matrix}\right.\)
câu c này mk sữa đề nhát
