Rút gọn các biểu thức sau: \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}\)
Rút gọn các biểu thức sau: \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}\)
Rút gọn các biểu thức sau: A=\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}\)
\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}-1}-1\right)}{\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}+2\right)}{\sqrt{3}}=\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}+2\)
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: (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) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
Rút gọn biểu thức sau :
\(N=1+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3+\sqrt{8}}}+...+\dfrac{1}{\sqrt{2017+\sqrt{2017^2-1}}}\)
Rút gọn các biểu thức sau:
a) $E=2 \sqrt{40 \sqrt{12}}+3 \sqrt{5 \sqrt{48}}-2 \sqrt{\sqrt{75}}-4 \sqrt{15 \sqrt{27}}$ :
b) $F=\dfrac{1}{\sqrt{3}}+\dfrac{1}{3 \sqrt{2}}+\dfrac{1}{\sqrt{3}} \sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}} .$
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
Rút gọn các biểu thức sau
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\) b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\) c) \(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\) e) \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{12-3\sqrt{7}}-\sqrt{2}\cdot\sqrt{12+3\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}\right)^2-2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}\right)^2+2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}\)
\(=-\sqrt{6}\)
c) \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{3\cdot9}{4\cdot16}}\)
\(=\sqrt[3]{\left(\dfrac{3}{4}\right)^3}\)
\(=\dfrac{3}{4}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)
\(=\sqrt[3]{\dfrac{54}{-2}}\)
\(=\sqrt[3]{-27}\)
\(=\sqrt[3]{\left(-3\right)^3}\)
\(=-3\)
a: Sửa đề: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}\cdot\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{2\sqrt{2}\left(\sqrt{6}+1\right)+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)
\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)
e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{2\sqrt{2}+3\sqrt{2}+6+1}-\sqrt[3]{2\sqrt{2}-3\sqrt{2}+6-1}\)
\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)
\(=\sqrt{2}+1-\left(\sqrt{2}-1\right)\)
\(=\sqrt{2}+1-\sqrt{2}+1=2\)
Bài 1 : (2 điểm) : Thực hiện phép tính và rút gọn các biểu thức sau :
a)A\(=-\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\sqrt{5}\)
b)\(B=\sqrt{48+\sqrt{5\dfrac{1}{3}+2\sqrt{75}-\sqrt[5]{1\dfrac{1}{3}}}}\)
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
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: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)