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bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
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\(\dfrac{2}{1-\sqrt{2}}-\dfrac{2}{1+\sqrt{2}}\)
\(=\dfrac{2\left(1+\sqrt{2}\right)-2\left(1-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}\)
\(=\dfrac{2+2\sqrt{2}-2+2\sqrt{2}}{1-2}=-4\sqrt{2}\)
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\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left[-\dfrac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-3\)
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\(\dfrac{2}{7+4\sqrt{3}}+\dfrac{2}{7-4\sqrt{3}}\)
\(=\dfrac{2\left(7-4\sqrt{3}\right)+2\left(7+4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\dfrac{14-8\sqrt{3}+14+8\sqrt{3}}{49-48}\)
= 28
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\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{4}{6-2\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}-\dfrac{2}{\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)-2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2\sqrt{5}-2-2\sqrt{5}-2}{5-1}\)
= - 1
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\(\dfrac{4}{1-\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
\(=\dfrac{4\left(1+\sqrt{3}\right)}{1-3}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)}\)
\(=-2-2\sqrt{3}-\sqrt{3}=-2-3\sqrt{3}\)
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\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\) (nhân [căn 2] vào cả tử và mẫu)
\(=\dfrac{2}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}\)
\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{2\left(5-\sqrt{5}\right)}{25-5}=\dfrac{5-\sqrt{5}}{10}\)
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
1/
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-\dfrac{4-3}{2-\sqrt{3}}\)
\(=\sqrt{3}+2+\sqrt{2}-\dfrac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{2-\sqrt{3}}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
2/
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\dfrac{\left(\sqrt{5}\right)^2}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\dfrac{\left(\sqrt{5}\right)^2}{\sqrt{5}}-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(\sqrt{5}+\sqrt{2}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=-\left(5-2\right)=-3\)
#F.C
b: \(B=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{5}+\dfrac{1}{2}\cdot2\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
Câu a, b, bạn có thể làm được suy nghĩ đi nha
c)
Ta có pt tổng quát :
\(\dfrac{1}{a\sqrt{a+1}+\left(a+1\right)\sqrt{a}}=\dfrac{1}{\sqrt{a\left(a+1\right)}\left(\sqrt{a}+\sqrt{\left(a+1\right)}\right)}=\dfrac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{a+1}}\)\(\Rightarrow C=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)..........Kaito Kid.......
\(A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(B=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\)
\(C=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(7-5\right)=2\sqrt{9}\)
\(D=\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(E=\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{5^2-\sqrt{5}^2}=\dfrac{60}{20}=3\)
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) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
sửa đề \(\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-\dfrac{11\left(2\sqrt{3}-3\right)}{3}\)
\(=\sqrt{5}+\sqrt{5}-1-\dfrac{11\left(2\sqrt{3}-3\right)}{3}=2\sqrt{5}-1-\dfrac{11\left(2\sqrt{3}-3\right)}{3}\)
\(=\dfrac{6\sqrt{5}-3-22\sqrt{3}+33}{3}=\dfrac{30+6\sqrt{5}-22\sqrt{3}}{3}\)
Sao lại sửa đề vậy ạ mình ghi đúng đề mà=((