\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
tinh ho minh voi, thanks
Những câu hỏi liên quan
\(P=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a)Rut gon P?
b)Tinh gia tri cua P voi \(x=3-2\sqrt{2}\)?
\(\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}}{2-\sqrt{6}}+\frac{\sqrt{3}}{2+\sqrt{6}}\right)-\frac{1}{\sqrt{2}}\)
mn giai nhanh ho mk nha ai nhanh nhat mk cho 3 k
Thuc hien phep tinh:
B=\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
tinh
a. sqrt{5}-sqrt{48}+5sqrt{27}-sqrt{45}
b.left(sqrt{5}+sqrt{2}right)left(3sqrt{2}-1right)
c.3sqrt{50}-2sqrt{75}-4frac{sqrt{54}}{sqrt{3}}-3sqrt{frac{1}{3}}
d.sqrt{left(sqrt{3}-3right)^2}+sqrt{4-2sqrt{3}}
e.frac{5sqrt{2}-2sqrt{5}}{sqrt{5}-sqrt{2}}+frac{6}{2-sqrt{10}}-frac{20}{sqrt{10}}
f.frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}+frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}+1}{sqrt{5}-1}
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tinh
a. \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)
b.\(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)
c.\(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)
d.\(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
e.\(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)
f.\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
\(\frac{\sqrt{13+2\sqrt{11}}+\sqrt{13-2\sqrt{11}}}{\sqrt{13+5\sqrt{5}}}-\sqrt{3-2\sqrt{2}}\)
Giup minh voi
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
Hay cho minh thay canh tay cua may ban gioi toan naooooo
Tinh gia tri bieu thuc
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}.\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{1+\sqrt{\frac{2+\sqrt{3}}{2}}}\)\(+\frac{\frac{2-\sqrt{3}}{2}}{1-\sqrt{\frac{2-\sqrt{3}}{2}}}\)
\(=\frac{\frac{4+2\sqrt{3}}{4}}{1+\sqrt{\frac{4+\sqrt{3}}{4}}}\)\(+\frac{\frac{4-2\sqrt{3}}{4}}{1-\sqrt{\frac{4-2\sqrt{3}}{4}}}\)
\(=\frac{\frac{3+2\sqrt{3}+1}{4}}{1+\sqrt{\frac{3+2\sqrt{3}+1}{4}}}\)\(+\frac{\frac{3-2\sqrt{3}+1}{4}}{1-\sqrt{\frac{3-2\sqrt{3}+1}{4}}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1+\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{1+\frac{\sqrt{3}+1}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{1-\frac{\sqrt{3}-1}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{2+\sqrt{3}}{2}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{2-\sqrt{3}}{2}}\)
\(=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{\left(\sqrt{3}+1\right)^2}{4}}\)\(+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{\left(\sqrt{3}-1\right)^2}{4}}\)
\(=1+1=2\)
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\(A=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(A=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(1-\frac{\sqrt{3}}{2}\right)}{2-\sqrt{4-2\sqrt{3}}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(A=\frac{\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\)
\(A=\frac{3+\sqrt{3}+3-\sqrt{3}}{6}\)
\(A=\frac{6}{6}=1\)
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Chứng minh:
\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
có sai đề hk z??/ bạn xem lại đáp án đi, mình làm ra kết quả khác ah
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Đúng mà bạn, nhưng mà từ để coi lại cái đã ^^
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\(VT\)\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\) \(+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+\sqrt{3}}}\)\(+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-\sqrt{3}}}\).......................
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Tính:a) frac{2}{3+2sqrt{2}}-frac{7}{1-2sqrt{2}}+frac{4}{sqrt{5}-1}+sqrt{8}-2b) frac{1}{1+sqrt{2}}+frac{1}{sqrt{3}+sqrt{2}}+frac{1}{sqrt{4}+sqrt{5}}c) sqrt{4-2sqrt{3}}+2sqrt{3}d) Asqrt{2-sqrt{3}}+sqrt{2+sqrt{3}}e) Bsqrt{frac{2}{2+sqrt{3}}}Các bạn giải đc câu nào thì giải dùm mk nha!!! Thanks :))))
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Tính:
a) \(\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)
b) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)
c) \(\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)
d) \(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
e) \(B=\sqrt{\frac{2}{2+\sqrt{3}}}\)
Các bạn giải đc câu nào thì giải dùm mk nha!!! Thanks :))))
\(a,\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)
\(=\frac{2.\left(3-2\sqrt{2}\right)}{9-8}-\frac{7.\left(1+2\sqrt{2}\right)}{1-8}+\frac{4.\left(\sqrt{5}+1\right)}{5-1}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}-\frac{7.\left(1+2\sqrt{2}\right)}{-7}+\frac{4.\left(\sqrt{5}+1\right)}{4}+2\sqrt{2}-2\)
\(=6-4\sqrt{2}+1+2\sqrt{2}+\sqrt{5}+1+2\sqrt{2}-2\)
\(=6+\sqrt{5}\)
\(b,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)
\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\)
\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\)
\(=-1+\sqrt{2}+\sqrt{3}-\sqrt{2}-2+\sqrt{5}\)
\(=-3+\sqrt{3}+\sqrt{5}\)
\(c,\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{3}\)
\(=\sqrt{3}-1+2\sqrt{3}\)
\(=-1+3\sqrt{3}\)
\(d,A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2\sqrt{3}}{\sqrt{2}}\)
\(=\sqrt{6}\)
\(e,B=\sqrt{\frac{2}{2+\sqrt{3}}}\)
Ta có \(\frac{2}{2+\sqrt{3}}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}\)
Thay lại ta được \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
.... Đúng thì ủng hộ nha ....
Kết bạn với mình ... ;) ;)
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1) Chứng minh: \(2\sqrt{n}-3< \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}-2\forall n\ge2\)
2) Thu gọn: \(A=5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)