\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)=\(\sqrt{3^2-2.3\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)

=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

57.\(\sqrt{8-\sqrt{55}}=\sqrt{\dfrac{16-2.\sqrt{5}.\sqrt{11}}{2}}=\sqrt{\dfrac{\sqrt{11}^2-2.\sqrt{5}.\sqrt{11}+\left(\sqrt{5}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}-\sqrt{5}\right)^2}{2}}=\dfrac{\left|\sqrt{11}-\sqrt{5}\right|}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{5}}{\sqrt{2}}\)

58. \(\sqrt{7+\sqrt{33}}=\sqrt{\dfrac{14+2\sqrt{3}.\sqrt{11}}{2}}=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2\sqrt{3}.\sqrt{11}+\left(\sqrt{3}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{11}+\sqrt{3}\right)^2}{2}}=\dfrac{\left|\sqrt{11}+\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{3}}{\sqrt{2}}\)

mấy câu dưới bạn cũng làm tương tự thôi

60) \(\sqrt{7-3\sqrt{5}}=\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}=\dfrac{3\sqrt{2}-\sqrt{10}}{2}\)

61) \(\sqrt{23+3\sqrt{5}}=\dfrac{\sqrt{46+6\sqrt{5}}}{\sqrt{2}}=\dfrac{3\sqrt{5}+1}{\sqrt{2}}=\dfrac{3\sqrt{10}+\sqrt{2}}{2}\)

62) \(\sqrt{7-\sqrt{33}}=\dfrac{\sqrt{14-2\sqrt{33}}}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{22}-\sqrt{6}}{2}\)

63) \(\sqrt{8+\sqrt{55}}=\dfrac{\sqrt{16+2\sqrt{55}}}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{22}+\sqrt{10}}{2}\)

60) \(\sqrt{7-3\sqrt{5}}=\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}=\dfrac{\left(3-\sqrt{5}\right)}{\sqrt{2}}=\dfrac{3\sqrt{2}-\sqrt{10}}{2}\)

59) \(\sqrt{6+\sqrt{35}}=\dfrac{\sqrt{12+2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{14}+\sqrt{10}}{2}\)

61) \(\sqrt{23+3\sqrt{5}}=\dfrac{\sqrt{46+6\sqrt{5}}}{\sqrt{2}}=\dfrac{3\sqrt{5}+1}{\sqrt{2}}=\dfrac{3\sqrt{10}+\sqrt{2}}{2}\)

62) \(\sqrt{7-\sqrt{33}}=\dfrac{\sqrt{14-2\sqrt{33}}}{\sqrt{2}}=\dfrac{\sqrt{11}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{22}-\sqrt{6}}{2}\)

63) \(\sqrt{8+\sqrt{55}}=\dfrac{\sqrt{16+2\sqrt{55}}}{\sqrt{2}}=\dfrac{\sqrt{11}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{22}+\sqrt{10}}{2}\)

b: Δ=(-12)^2-4*2*(9+4căn 2)

=144-72-32căn 2=72-32căn 2

=(8-2căn 2)^2

=>PT có hai nghiệm pb là:

\(\left\{{}\begin{matrix}x=\dfrac{12-8+2\sqrt{2}}{4}=\dfrac{2+\sqrt{2}}{2}\\x_2=\dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\)

c: Δ=(-30)^2-4*3*(-26+8căn 3)

=900+312-96căn 3

=1212-2*căn 3072

=>Phương trình có hai nghiệm pb là:

\(\left\{{}\begin{matrix}x=\dfrac{30-2\sqrt{1212-2\sqrt{3072}}}{6}\\x=\dfrac{30+2\sqrt{1212-2\sqrt{3072}}}{6}\end{matrix}\right.\)

\(2\sqrt{3}\left(\sqrt{12}-\sqrt{27}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=2\sqrt{3}\left(\sqrt{3.2^2}-\sqrt{3.3^2}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=2\sqrt{3}\left(2\sqrt{3}-3\sqrt{3}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=\left(2\sqrt{3}\right)^2-6.\left(\sqrt{3}\right)^2-6\sqrt{6}+6\sqrt{6}\)

\(=-6\)

vt phép tính rõ ra đi

vt là viết

1)\(x\le\dfrac{-1}{2}\)

2)\(x^2+3\ge3\forall x\)

3)\(-2\le x\le2\)