Bài 1: Căn bậc hai
Các câu hỏi tương tự
1/Tính
Adfrac{sqrt{15-10sqrt{2}}+sqrt{13+4sqrt{10}}-sqrt{11+2sqrt{10}}}{2sqrt{3+2sqrt{2}}+sqrt{9-4sqrt{2}}+sqrt{12+8sqrt{2}}}
Bdfrac{2+sqrt{3}}{sqrt{2}+sqrt{2}+sqrt{3}}+dfrac{2-sqrt{3}}{sqrt{2}-sqrt{2}-sqrt{3}}
Cdfrac{sqrt{2-sqrt{3}}+sqrt{4-sqrt{15}}+sqrt{10}}{sqrt{23-3sqrt{5}}}
Ddfrac{sqrt{4+sqrt{3}}+sqrt{4-sqrt{3}}}{sqrt{4+sqrt{13}}}
2/So sánh
sqrt{2017^2-1}-sqrt{2016^2-1} và dfrac{2.1016}{sqrt{2017^2-1}+sqrt{2016^2-1}}
Đọc tiếp
1/Tính
A=\(\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
B=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)
C=\(\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
D=\(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}\)
2/So sánh
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}\) và \(\dfrac{2.1016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Tính tổng sau: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
So sánh A=\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)
với B=\(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)
Rút gọn:1) dfrac{1}{sqrt{3}+1}+dfrac{1}{sqrt{3}-1}-2sqrt{3}Pleft(dfrac{1}{sqrt{x}-1}-dfrac{1}{sqrt{x}}right):left(dfrac{sqrt{x}+1}{sqrt{x}-2}-dfrac{sqrt{x}+2}{sqrt{x}-1}right)2) sqrt{3-2sqrt{2}}+dfrac{1}{sqrt{2}-1}Mleft(dfrac{sqrt{a}}{sqrt{a}-2}+dfrac{sqrt{a}}{sqrt{a}+2}right).dfrac{a-4}{sqrt{4a}}Nleft(1-dfrac{sqrt{x}}{sqrt{x}+1}right):left(dfrac{sqrt{x}+2}{sqrt{x}+3}+dfrac{sqrt{x}-3}{2-sqrt{x}}+dfrac{sqrt{x}-2}{x+sqrt{x}-6}right)Qleft(1-dfrac{sqrt{x}}{sqrt{x}+1}right):left(dfrac{sqrt{x}+2}{sqr...
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Rút gọn:
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right).\dfrac{a-4}{\sqrt{4a}}\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
Làm chi tiết giúp mình với vì mình yếu phần này lắm
Tính: \(A=\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
thực hiện phép tính:
1, dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}-dfrac{5-2sqrt{5}}{2sqrt{5}-4}
2,dfrac{sqrt{3}+1}{sqrt{3}-1}+dfrac{sqrt{3}-1}{sqrt{3}+1}
3,dfrac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}+dfrac{sqrt{5}+sqrt{27}}{sqrt{30}-sqrt{2}}
4,dfrac{3-sqrt{3}}{2sqrt{3}-1}+dfrac{3+sqrt{3}}{2sqrt{3}-1}
5,dfrac{2sqrt{3}-4}{sqrt{3}-1}+dfrac{2sqrt{2}-1}{sqrt{2}-1}-dfrac{1+sqrt{6}}{sqrt{2}+sqrt{3}}
Đọc tiếp
thực hiện phép tính:
1, \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2,\(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}+\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
4,\(\dfrac{3-\sqrt{3}}{2\sqrt{3}-1}+\dfrac{3+\sqrt{3}}{2\sqrt{3}-1}\)
5,\(\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
10 tháng 7 2023 lúc 11:17
\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3-\sqrt{2}}{3+\sqrt{2}}\)
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
Tính: S = \(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}\) + \(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}\) + \(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}\) + ... + \(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)
Mong các bạn giúp giùm với ak, mk cảm ơn nhìu lắm
Tìm số tự nhiên n , biết rằng : \(\dfrac{1}{\sqrt{1^3+2^3}}+\dfrac{1}{\sqrt{1^3+2^3+3^3}}+...+\dfrac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\dfrac{2015}{2017}\)