\(A=2\sqrt{6}\)

\(B=2\sqrt{4}=4\)

\(C=2\sqrt{7}\)

Bn lm rõ ra chút đc k ak

a/ \(\left(\sqrt{18}\right)^2-2\cdot\sqrt{18}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{18}-\sqrt{3}\right)^2\)

b/\(\left(\sqrt{54}\right)^2-2\cdot\sqrt{54}+1=\left(\sqrt{54}-1\right)^2\)

c/\(\left(\sqrt{9}\right)^2-2\cdot\sqrt{9}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{9}-\sqrt{5}\right)^2\)

d/\(\left(\sqrt{8}\right)^2+2\cdot\sqrt{8}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{8}+\sqrt{5}\right)^2\)

ĐKXĐ: x>=0

Ta có: \(3\sqrt{2x}-2\sqrt2=\sqrt6-\sqrt{3x}\)

=>\(3\sqrt{2x}+\sqrt{3x}=2\sqrt2+\sqrt6\)

=>\(\sqrt{18x}+\sqrt{3x}=2\sqrt2+\sqrt6\)

=>\(\sqrt{x}\left(\sqrt{18}+\sqrt3\right)=2\sqrt2+\sqrt6\)

=>\(\sqrt{x}=\frac{2\sqrt2+\sqrt6}{\sqrt{18}+\sqrt3}=\frac{\sqrt8+\sqrt6}{\sqrt{18}+\sqrt3}=\frac{\left(\sqrt8+\sqrt6\right)\left(\sqrt{18}-\sqrt3\right)}{18-3}=\frac{\sqrt{144}-\sqrt{24}+\sqrt{108}-\sqrt{18}}{15}=\frac{12-2\sqrt6+6\sqrt3-3\sqrt2}{15}\)

=>\(x=\frac{\left(12-2\sqrt6+6\sqrt3-3\sqrt2\right)^2}{225}\) (nhận)

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

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

à cộng \(\sqrt{2}\)bạn nhé

\(E=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(4+\sqrt{15}\right)^2}.\sqrt{\left(\sqrt{10}-\sqrt{6}\right)^2}.\frac{4^2-15}{\sqrt{4+\sqrt{15}}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{10+6-2\sqrt{10}.\sqrt{6}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{16-2\sqrt{60}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{4\left(4-\sqrt{15}\right)}\)

\(=2\sqrt{\left(4+\sqrt{15}\right).\left(4-\sqrt{15}\right)}\)

\(=2\sqrt{16-15}=2\)

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

Trả lời:

\(A=\sqrt{3}-\frac{\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

\(A=\sqrt{3}+\frac{\sqrt{6}}{\sqrt{2}-1}-\frac{2\sqrt{2}+2}{\sqrt{2}+1}\)

\(A=\sqrt{3}+\frac{\sqrt{6}.\left(\sqrt{2}+1\right)}{2-1}-\frac{2.\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)

\(A=\sqrt{3}+\sqrt{6}.\left(\sqrt{2}+1\right)-2\)

\(A=\sqrt{3}+\sqrt{12}+\sqrt{6}-2\)

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

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

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