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\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+...+\frac{1}{\sqrt{23}+\sqrt{25}}\)
\(2A=\frac{2}{\sqrt{3}+\sqrt{1}}+\frac{2}{\sqrt{5}+\sqrt{3}}+...+\frac{2}{\sqrt{25}+\sqrt{23}}\)\(2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{\left(\sqrt{3}+\sqrt{1}\right)\left(\sqrt{3}-\sqrt{1}\right)}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{\left(\sqrt{25}+\sqrt{23}\right)\left(\sqrt{25}-\sqrt{23}\right)}\)
\(2A=\frac{2\left(\sqrt{3}-\sqrt{1}\right)}{2}+\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}+...+\frac{2\left(\sqrt{25}-\sqrt{23}\right)}{2}\)
\(2A=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{25}-\sqrt{23}\)
\(2A=\sqrt{25}-\sqrt{1}\)
\(2A=4\)
\(A=2\)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
a ) \(\dfrac{2}{\sqrt{3}-1}\) - \(\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
= \(\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\) = \(\dfrac{4}{2}\) = 2
b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)
= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)
c) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\) = \(\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
\(a,=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\\ b,=\sqrt{\left(3-\sqrt{5}\right)^2}+\left|2-\sqrt{5}\right|=3-\sqrt{5}+\sqrt{5}-2=1\\ c,=\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{2}-\dfrac{-\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=\sqrt{5}-\sqrt{3}+\sqrt{3}=\sqrt{5}\)
\(A=\dfrac{\sqrt{6+2\sqrt{5}}}{2-\sqrt{6-2\sqrt{5}}}-\dfrac{\sqrt{6-2\sqrt{5}}}{2+\sqrt{6+2\sqrt{5}}}\)
\(=\dfrac{\sqrt{5}+1}{2-\sqrt{5}+1}-\dfrac{\sqrt{5}-1}{3+\sqrt{5}}\)
\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)
\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)
\(=4\)
b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)
\(\dfrac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{-2-6\sqrt{15}}{5-3}=\dfrac{-2\left(1+3\sqrt{15}\right)}{2}=-\left(1+3\sqrt{15}\right)=-1-3\sqrt{15}\)
nhân tử vs mẫu vs \(\sqrt{5}-\sqrt{3}\)
và nhân tử vs mẫu vs\(\sqrt{5}-\sqrt{3}\)
\(=\dfrac{8+2\sqrt{15}+8-2\sqrt{15}}{2}\)
=8
nhân liên hợp lên là ra nha bạn! ('ω')