Mình ko hiểu đề bạn giải thích rõ hơn được ko

\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\Leftrightarrow A^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{9-5}\)

\(A^2=6+4=10\Rightarrow A=\sqrt{10}\)

\(3\sqrt{8}-\sqrt{50}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=6\sqrt{2}-5\sqrt{2}-\left(\sqrt{2}-1\right)\)

\(=\sqrt{2}-\sqrt{2}+1\)

\(=1\)

Phép 1:

Ta có: \(3\cdot\sqrt{7-4\sqrt{3}}\)

\(=3\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=3\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=3\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)

Phép 2:

Ta có: \(\sqrt{11+4\sqrt{7}}\)

\(=\sqrt{7+2\cdot\sqrt{7}\cdot2+4}\)

\(=\sqrt{\left(\sqrt{7}+2\right)^2}\)

\(=\left|\sqrt{7}+2\right|\)

\(=\sqrt{7}+2\)(Vì \(\sqrt{7}+2>0\))

Phép 3:

Ta có: \(2\cdot\sqrt{11-4\sqrt{7}}\)

\(=2\cdot\sqrt{7-2\cdot\sqrt{7}\cdot2+4}\)

\(=2\cdot\sqrt{\left(\sqrt{7}-2\right)^2}\)

\(=2\cdot\left|\sqrt{7}-2\right|\)

\(=2\cdot\left(\sqrt{7}-2\right)\)(Vì \(\sqrt{7}>2\))

\(=2\sqrt{7}-4\)

Phép 4:

Ta có: \(\sqrt{19-4\sqrt{15}}\)

\(=\sqrt{15-2\cdot\sqrt{15}\cdot2+4}\)

\(=\sqrt{\left(\sqrt{15}-2\right)^2}\)

\(=\left|\sqrt{15}-2\right|\)

\(=\sqrt{15}-2\)(Vì \(\sqrt{15}>2\))

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