Đặt \(A=\sqrt{24+12\sqrt{3}}\)
\(\Rightarrow\sqrt{\frac{1}{2}}A=\sqrt{\frac{1}{2}}.\sqrt{24+12\sqrt{3}}\)
\(\Leftrightarrow\frac{A}{\sqrt{2}}=\sqrt{12+6\sqrt{3}}\)
\(\Leftrightarrow\frac{A}{\sqrt{2}}=\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(\Leftrightarrow\frac{A}{\sqrt{2}}=3+\sqrt{3}\)
\(\Leftrightarrow A=3\sqrt{2}+\sqrt{6}\)
\(\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{4}{1-\sqrt{7}}\)
\(\sqrt{24+12\sqrt{3}}\)
bài 1 rút gọn biểu thức sau:
a)\(\sqrt{16+6\sqrt{7}}\)- \(\sqrt{8-2\sqrt{7}}\) b)K=\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
c)\(\sqrt{60-24\sqrt{6}}\)+\(\sqrt{40-16\sqrt{6}}\) d)B=(3+\(\sqrt{3}\))\(\sqrt{12-6\sqrt{13}}\)
e)\(\sqrt{6-4\sqrt{2}}\)-\(\sqrt{\left(\sqrt{2}-\sqrt{6}\right)^2}\)
bài 2 cho biểu thức A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{x}+3}{x+9}\)( với x≥0 và x≠ 9)
a) rút gọn biểu thức A
b) tính giá trị biểu thức\(x=4+2\sqrt{3}\)
Rg
a) \(\dfrac{(3\sqrt{3}+\sqrt{50})(5-\sqrt{24})}{\sqrt{75}-5\sqrt{2}}\)
b)\((\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}})(\sqrt{6}+11)\)
\(\sqrt{24-16\sqrt{2}}+\sqrt{12-8\sqrt{2}}\)
tính giá trị biểu thức
\(\sqrt{17-12\sqrt{2}}\) -\(\sqrt{24-8\sqrt{8}}\)
Rút gọn các biểu thức sau:
9, A = \(\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\)
10, A = \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
11, A = \(\text{}\text{}\text{}\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
12, A = \(\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-3\sqrt{3}}\)
13, A = \(\sqrt{9-4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
A=\(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)
B=\(5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\dfrac{3}{2}}\right)^2\)
