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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
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a) \(\frac{\left(\sqrt{3}-\sqrt{5}\right)^2+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}-\sqrt{5}\)
= \(\frac{3-2\sqrt{15}+5+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}-\sqrt{5}\)
=\(\frac{8+2\sqrt{15}-\sqrt{5}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}\)
= \(\frac{8+2\sqrt{15}-\sqrt{15}-\sqrt{25}}{\sqrt{3}+\sqrt{5}}\)
= \(\frac{3+\sqrt{15}}{\sqrt{3}+\sqrt{5}}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}\)
= \(\sqrt{3}\)
b) \(\frac{\left(\sqrt{2}+1\right)^2-4\sqrt{2}}{\sqrt{2}-1}.\left(\sqrt{2}+1\right)\)
= \(\frac{2+2\sqrt{2}+1-4\sqrt{2}}{\sqrt{2}-1}.\left(\sqrt{2}+1\right)\)
= \(\frac{\left(\sqrt{2}-1\right)^2.\left(\sqrt{2}+1\right)}{\sqrt{2}-1}\)
= \(\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)\)
= 2 - 1
= 1