\(...=\left(\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-4\dfrac{\sqrt{6}}{2}\right)\left(\dfrac{3\sqrt{6}}{3}-2\sqrt{3}-\sqrt{6}\right)\) (Trục căn)
\(=\left(\dfrac{9\sqrt{6}}{6}+\dfrac{4\sqrt{6}}{6}-\dfrac{12\sqrt{6}}{6}\right)\left(\sqrt{6}-2\sqrt{3}-\sqrt{6}\right)\)
\(=\dfrac{\sqrt{6}}{6}.\left(-2\sqrt{3}\right)=\dfrac{1}{\sqrt{6}}.\left(-2\sqrt{3}\right)=-\sqrt{2}\)
RÚT GỌN BIỂU THỨC
A= \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\)\(\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
B= \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\)\(\left(\sqrt{6}+11\right)\)
1)\(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
2)\(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
3)\(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
Trục căn thức ở mẫu và rút gọn
a)\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
b)\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Help me plssssssss
Rút gọn
\(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{3}{2}+2}\right)\left(\dfrac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\dfrac{\sqrt{3}}{2+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\dfrac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
Chứng minh các đẳng thức sau:
e) \(\left(\dfrac{3}{2}.\sqrt{6}+2.\sqrt{\dfrac{2}{3}}-4.\sqrt{\dfrac{3}{2}}\right).\left(\dfrac{3}{2}.\sqrt{6}+2.\sqrt{\dfrac{2}{3}}+4.\sqrt{\dfrac{3}{2}}\right)=-\sqrt{2}\)
a) A=\(\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{\sqrt{15}-\sqrt{35}}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
b) B=\(\dfrac{12}{3+\sqrt{3}}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{27}-3\sqrt{2}}{\sqrt{3}.\sqrt{2}}\)
c)C=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)(x>0,x≠1,x≠4)
Rút gọn:
a) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
b) \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
c) \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)
d) \(\sqrt{\left(\sqrt{3}+4\right)\sqrt{19-8\sqrt{3}}+3}\)
e) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
Tính
a)\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{ }6}\)
b) \(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)
thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{3\sqrt{6}}{\sqrt{2}}+\dfrac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
b) \(\left(\dfrac{2-2\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
