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\(1.\dfrac{\sqrt{30}-\sqrt{2}}{\sqrt{8}-\sqrt{15}}-\sqrt{8-\sqrt{49+8\sqrt{3}}}=\dfrac{\sqrt{60}-\sqrt{4}}{\sqrt{16-2\sqrt{15}}}-\sqrt{8-\sqrt{48+2.4\sqrt{3}+1}}=\dfrac{2\left(\sqrt{15}-1\right)}{\sqrt{\left(\sqrt{15}-1\right)^2}}-\sqrt{8-|4\sqrt{3}+1|}=2-\sqrt{4-2.2\sqrt{3}+3}=2-|2-\sqrt{3}|=\sqrt{3}\)
\(2.\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{\sqrt{4}+\sqrt{4+2\sqrt{3}}}+\dfrac{2\sqrt{2}-\sqrt{6}}{\sqrt{4}-\sqrt{4-2\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+|\sqrt{3}+1|}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-|\sqrt{3}-1|}=\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}=\dfrac{12\sqrt{2}-2\sqrt{18}}{9-3}=\dfrac{12\sqrt{2}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
\(3.\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{2}{4+\sqrt{5+2\sqrt{5}+1}}+\dfrac{2}{4-\sqrt{5-2\sqrt{5}+1}}=\dfrac{2}{4+|\sqrt{5}+1|}+\dfrac{2}{4-|\sqrt{5}-1|}=\dfrac{2}{\sqrt{5}+5}+\dfrac{2}{5-\sqrt{5}}=\dfrac{10-2\sqrt{5}+10+2\sqrt{5}}{20}=\dfrac{20}{20}=1\)
a. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+14\sqrt{2}=14-14\sqrt{2}+7+14\sqrt{2}=21\)
b. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}-\dfrac{\sqrt{5}}{2}=\dfrac{2\sqrt{5}-\sqrt{5}}{2}=\dfrac{\sqrt{5}}{2}\)
c. \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+2\sqrt{7}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
Kkk
a) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=1\)
b)
\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2\sqrt{6}}{6}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{6+3\sqrt{3}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}}{6\sqrt{3}}}\)
\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{6+3\sqrt{3}}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\left(-\sqrt{3}+2+\sqrt{3}\right)}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{12+6\sqrt{3}}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}}{-2\sqrt{3}}\)
\(=\dfrac{3+\sqrt{3}}{-2\sqrt{3}}\)
\(=-\dfrac{\left(3+\sqrt{3}\right)\sqrt{3}}{6}\)
\(=-\dfrac{3\sqrt{3}+3}{6}\)
\(=-\dfrac{3\left(\sqrt{3}+3\right)}{6}\)
\(=-\dfrac{\sqrt{3}+1}{2}\)
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
Bài 1: Ta có:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{6-2}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{6+2-2\sqrt{6.2}}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{(\sqrt{6}-\sqrt{2})^2}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2}=\frac{6-2}{2}=2\)
Bài 2:
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)
\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)
\(=16+2\sqrt{64-4(10+2\sqrt{5})}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)
\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}\)
\(=16+2(\sqrt{20}-2)=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)
\(\Rightarrow A=\sqrt{10}+\sqrt{2}\)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
a,\(\sqrt{2\left(11+6\sqrt{2}\right)}\)=\(\sqrt{2\left(9+2.3.\sqrt{2}+2\right)}\)=\(\sqrt{2\left(3+\sqrt{2}\right)^2}\)=\(\sqrt{2}\)(3+\(\sqrt{2}\))
\(a.\sqrt{22+12\sqrt{2}}=\sqrt{18+2.3\sqrt{2}.2+4}=3\sqrt{2}+2\)
\(b.\sqrt{\dfrac{5+2\sqrt{6}}{2}}=\sqrt{\dfrac{3+2\sqrt{3}.\sqrt{2}+2}{2}}=\dfrac{\sqrt{3}+\sqrt{2}}{2}\)
\(c.\sqrt{30+4\sqrt{2}.\sqrt{7}}=\sqrt{28+2.\sqrt{2}.2\sqrt{7}+2}=2\sqrt{7}+\sqrt{2}\)
\(d.\sqrt{5+2\sqrt{2-\sqrt{9-4\sqrt{2}}}}=\sqrt{5+2\sqrt{2-\sqrt{8-2.2\sqrt{2}+1}}}=\sqrt{5+2\sqrt{2-2\sqrt{2}+1}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{2}+1\) \(e.\sqrt{1+2\sqrt{\sqrt{2+\sqrt{11+6\sqrt{2}}}}}=\sqrt{1+2\sqrt{\sqrt{2+\sqrt{9+2.3\sqrt{2}+2}}}}=\sqrt{1+2\sqrt{\sqrt{5+\sqrt{2}}}}\)
\(f.\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{1-\dfrac{\sqrt{3}}{2}}}=\sqrt{1+\dfrac{\sqrt{3}}{2}+\sqrt{\dfrac{3}{4}-2.\dfrac{\sqrt{3}}{2}.\dfrac{1}{2}+\dfrac{1}{4}}}=\sqrt{\sqrt{3}+\dfrac{1}{2}}=\)
\(g.\sqrt{10-2\sqrt{21}}+\sqrt{4+2\sqrt{3}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}+\sqrt{3+2\sqrt{3}+1}=\sqrt{7}-\sqrt{3}+\sqrt{3}+1=\sqrt{7}+1\)
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: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
bài này trước, bài trên để coi lại đã
Đặt \(A=\sqrt{2-\sqrt{3}}-\sqrt{\dfrac{3}{2}}\Rightarrow A\sqrt{2}=\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\Rightarrow A=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\)
đặt \(A=\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)
\(\Rightarrow A\sqrt{2}=\dfrac{\sqrt{12-2\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6\sqrt{2}}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}+1}=\dfrac{\sqrt{11}-1}{\sqrt{2}\left(\sqrt{11}-1\right)}-\dfrac{6\sqrt{2}}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}+1}=\dfrac{1}{\sqrt{2}}-\dfrac{6\sqrt{2}}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}+1}=\dfrac{\sqrt{2}+1-12-6\sqrt{2}-6}{\sqrt{2}\left(\sqrt{2}+1\right)}=\dfrac{-17-5\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)
\(\Rightarrow A=\dfrac{-17-5\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+1\right)}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{-17-5\sqrt{2}}{2\left(\sqrt{2}+1\right)}=\dfrac{\left(-17-5\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2}=\dfrac{7-12\sqrt{2}}{2}\)
Kl: \(A=\dfrac{7-12\sqrt{2}}{2}\)
\(1.\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}=\dfrac{\sqrt{12-2\sqrt{11}}}{\sqrt{4.11}-2}-\dfrac{6\sqrt{2}}{2}-\dfrac{3\left(\sqrt{2}-1\right)}{2-1}=\dfrac{\sqrt{\left(\sqrt{11}-1\right)^2}}{2\sqrt{11}-2}-3\sqrt{2}-3\sqrt{2}+3=\dfrac{\sqrt{11}-1}{2\left(\sqrt{11}-1\right)}+3=\dfrac{1}{2}+3=\dfrac{7}{2}\)
\(2.\sqrt{2-\sqrt{3}}-\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{\dfrac{6}{2}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}}{\sqrt{2}}=\dfrac{\text{ |}\sqrt{3}-1\text{ |}-\sqrt{3}}{\sqrt{2}}=\dfrac{-1}{\sqrt{2}}=\dfrac{-\sqrt{2}}{2}\)