Rút gọn biểu thức sau :
\(N=1+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3+\sqrt{8}}}+...+\dfrac{1}{\sqrt{2017+\sqrt{2017^2-1}}}\)
Bài 1: Rút gọn biểu thức sau:
a. \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}\)
b. \(B=\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{2\sqrt{3}+3\sqrt{2}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2018\sqrt{2017}+2017\sqrt{2018}}\)
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
b/ \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow B=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2017}}-\dfrac{1}{\sqrt{2018}}=1-\dfrac{1}{\sqrt{2018}}\)
rút gọn các biểu thức sau: (giả thiết các biểu thức chữ đều có nghĩa)
a) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
b) \(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\)
c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
Thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{\sqrt{2}}{2\sqrt{2}-3}\)+\(\dfrac{1}{3+2\sqrt{2}}\) b) \(\dfrac{1}{\sqrt{10}+\sqrt{6}}\)+\(\dfrac{1}{\sqrt{6}-\sqrt{10}}\)
c) \(\dfrac{-2}{3\sqrt{8}}\)+\(\dfrac{1}{3-2\sqrt{2}}\)
a: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}+3\right)+2\sqrt{2}-3}{-1}\)
\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)
b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)
\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)
c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)
\(=\dfrac{-2\left(3-2\sqrt{2}\right)+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{-6+4\sqrt{2}+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}\)
\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)
b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)
Rút gọn biểu thức sau:
1) \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
2) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{1}{1+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)
1) Ta có: \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\dfrac{10}{\sqrt{3}}\)
\(=\dfrac{-27+10}{\sqrt{3}}\)
\(=\dfrac{-17\sqrt{3}}{3}\)
b) Ta có: \(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{1}{\sqrt{2}+1}+\dfrac{\sqrt{2}+1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{2}-1-\sqrt{2}+3+2\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)
\(=\dfrac{2+2\sqrt{2}}{2+2\sqrt{2}}=1\)
Rút gọn các biểu thức :
A=\(\dfrac{1}{\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}}\)
B= \(\dfrac{1}{1+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2015}+\sqrt{2017}}\)
rút gọn các biểu thức sau
\(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)+\(\dfrac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
\(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\)+\(\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)
\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)
\(=\dfrac{3+2\sqrt{2}}{\sqrt{2}+1}+\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\)
=căn 2+1+căn 2-1=2căn 2
b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)
Rút gọn các biểu thức sau :
a) A= \(\sqrt{18}\) . \(\sqrt{2}\) - \(\sqrt{48}\) : \(\sqrt{3}\)
b)B= \(\dfrac{8}{\sqrt{5}-1}\) + \(\dfrac{8}{\sqrt{5}+1}\)
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
Rút gọn biểu thức sau:
\(\sqrt{12+6\sqrt{3}}-\sqrt{3}\)
\(\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\times\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)