https://i.imgur.com/agp1B8G.jpg
Bài 1: Căn bậc hai
Các câu hỏi tương tự
Rút gọn biểu thức:
Afrac{1}{sqrt{1}-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-frac{1}{sqrt{4}-sqrt{5}}
Bleft(frac{5-sqrt{5}}{sqrt{5}}-2right)left(frac{4}{1+sqrt{5}}+4right)
Cleft(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right):left(1:frac{1}{sqrt{2}+sqrt{3}}right)
D2sqrt{50}-frac{1}{sqrt{2}-1}+4sqrt{frac{9}{2}}-sqrt{3-2sqrt{2}}
Đọc tiếp
Rút gọn biểu thức:
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}\)
\(B=\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)
\(C=\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right):\left(1:\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(D=2\sqrt{50}-\frac{1}{\sqrt{2}-1}+4\sqrt{\frac{9}{2}}-\sqrt{3-2\sqrt{2}}\)
1.Chứng minh: \(\frac{1}{2\cdot\sqrt{1}}+\frac{1}{3\cdot\sqrt{2}}+\frac{1}{4\cdot\sqrt{3}}+...+\frac{1}{2012\cdot\sqrt{2011}}+\frac{1}{2013\cdot\sqrt{2012}}\)\(< 2\)
2.Chứng minh: A= \(\frac{1}{3\cdot\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\cdot\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\cdot\left(\sqrt{48}+\sqrt{49}\right)}\)\(< \frac{1}{2}\)
4 tháng 7 2019 lúc 20:01
Tính:
A= \(\frac{1}{1+\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{5}}}+\frac{1}{1+\sqrt{\frac{3}{5}}+\sqrt{\frac{3}{2}}}+\frac{1}{1+\sqrt{\frac{5}{2}}+\sqrt{\frac{5}{3}}}\)
CMR n\(\in\)N, n>3
a,\(\frac{1}{2\sqrt{1} }+\frac{1}{3\sqrt{2} } +\frac{1}{4\sqrt{3} }+...+\frac{1}{(n+1)\sqrt{n} }<2 \)
b,S=\(\frac{1}{3(1+\sqrt{2}) }+\frac{1}{5(\sqrt{2}+\sqrt{3} }+...+\frac{1}{(2n+1)(\sqrt{n}+\sqrt{n+1}) } \)
Chứng minh rằng \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}< 1\)
rút gọn:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
(1)frac{sqrt{6+4sqrt{2}}}{sqrt{2}} (2)frac{sqrt{3-sqrt{5}}}{sqrt{0.5}} (3)left(sqrt{2}-1right)^2 (4)left(3-2sqrt{2}right).left(3+2sqrt{2}right) (5)sqrt{left(2-sqrt{3}right)}^2-sqrt{left(1-sqrt{3}right)}^2 (6)sqrt{left(sqrt{2}-sqrt{3}right)^2}-sqrt{left(sqrt{2}+sqrt{3}right)}^2 (7)frac{1}{sqrt{3}-1}-frac{1}{sqrt{3}+1} (8)sqrt{3-2sqrt{2}} (9)sqrt{4+2sqrt{3}}-sqrt{4-2sqrt{3}} (10)sqrt{2020+2sqrt{2019}}-sqrt{2020-2sqrt{2019}} (11)sqrt{7+2sqrt{12}} Các bạn giúp mình với ,Mình xin cảm ơn trước
Đọc tiếp
(1)\(\frac{\sqrt{6+4\sqrt{2}}}{\sqrt{2}}\) (2)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{0.5}}\) (3)\(\left(\sqrt{2}-1\right)^2\) (4)\(\left(3-2\sqrt{2}\right).\left(3+2\sqrt{2}\right)\) (5)\(\sqrt{\left(2-\sqrt{3}\right)}^2-\sqrt{\left(1-\sqrt{3}\right)}^2\) (6)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)}^2\) (7)\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\) (8)\(\sqrt{3-2\sqrt{2}}\) (9)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (10)\(\sqrt{2020+2\sqrt{2019}}-\sqrt{2020-2\sqrt{2019}}\) (11)\(\sqrt{7+2\sqrt{12}}\) Các bạn giúp mình với ,Mình xin cảm ơn trước
a, Rút gọn A= \(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)
b, B= \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{49^2}+\frac{1}{50^2}}\)
Cho n là số tự nhiên khác 0. Tìm giá trị nhỏ nhất của
Q= \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+....+\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}+\frac{101}{n+1}\)