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\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)
\(=\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}\)
\(=14\)
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)
\(=1+\sqrt{2}\)
a) \(\frac{\sqrt{2.5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\left(\sqrt{2}\right)^2-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
(Hai bài náy đều tương tự nhau bạn ạ, nhớ k cho mình với nhé, chúc bạn học tốt!)
- \(\frac{2+\sqrt{2}}{1+\sqrt{2}}=\frac{\sqrt{2}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{2}\)
- \(\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=-\sqrt{5}\)
- \(\frac{2\sqrt{3}-\sqrt{6}}{1-\sqrt{3}}=\frac{-\sqrt{6}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=-\sqrt{6}\)
- \(\frac{a-\sqrt{a}}{1-\sqrt{a}}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
- \(\frac{p-2\sqrt{p}}{\sqrt{p}-2}=\frac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
a)
\(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}=3-2\sqrt{2}-4+\sqrt{8}\)
\(=3-2\sqrt{2}-4+2\sqrt{2}=3-4=-1\)
b)
\(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\frac{2\left(\sqrt{3}+1-\sqrt{3}+1\right)}{2}=\sqrt{3}+1-\sqrt{3}+1=1+1=2\)
2 cái đầu quy đồng r tính.... cái sau rút gọn cho căn 2