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B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
B=\(\sqrt{9+2.3\sqrt{6}+6}+\sqrt{9+2.3.2\sqrt{6}+24}=\sqrt{\left(3+\sqrt{6}\right)^2}+\sqrt{\left(3+2\sqrt{6}\right)^2}\)=\(=3+\sqrt{6}+2+2\sqrt{6}=5+3\sqrt{6}\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
a/ \(\sqrt{5+\sqrt{24}}-\sqrt{2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{2}=\left|\sqrt{3}+\sqrt{2}\right|-\sqrt{2}=\sqrt{3}+\sqrt{2}-\sqrt{2}=\sqrt{3}\)
b/ \(\frac{3-2\sqrt{3}}{\sqrt{3}-2}=\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}=\sqrt{3}\)
c/ \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
d/ \(\frac{1}{1-\sqrt{2}}-\frac{1}{1+\sqrt{2}}=\frac{1+\sqrt{2}-1+\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\frac{2\sqrt{2}}{1-2}=-2\sqrt{2}\)
22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)
\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)
\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}=\sqrt{6+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)
\(=\sqrt{2+3+1+2\sqrt{2}+2\sqrt{3}+2\sqrt{6}}\)\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}=\left|\sqrt{2}+\sqrt{3}+1\right|=\sqrt{2}+\sqrt{3}+1\)