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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) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
b, \(\frac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\) + \(\frac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
= \(\frac{\sqrt{3}\left(1+\sqrt{\sqrt{3}+1}\right)+\sqrt{3}\left(1-\sqrt{\sqrt{3}+1}\right)}{\left(1-\sqrt{\sqrt{3}+1}\right)\left(1+\sqrt{\sqrt{3}+1}\right)}\)
= \(\frac{\sqrt{3}+\sqrt{3\sqrt{3}+3}+\sqrt{3}-\sqrt{3\sqrt{3}+3}}{1^2-\left(\sqrt{\sqrt{3}+1}\right)^2}\)
= \(\frac{2\sqrt{3}}{1-\sqrt{3}-1}\)
= \(\frac{2\sqrt{3}}{-\sqrt{3}}\)
= -2
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