1.
biểu thức trên được xác định khi và chỉ khi :
\(5x^2-3x-8\ge0\\ x^2-\dfrac{3}{5}x-\dfrac{8}{5}\ge0\\ x^2-2.\dfrac{3}{10}+\dfrac{9}{100}\ge\dfrac{8}{5}+\dfrac{9}{100}\\ \left(x-0,3\right)^2\ge1,69\\ \Rightarrow x\ge1,6\)
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D= \(\dfrac{\left(5+4\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Rút gọn các biểu thức sau:
a) \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}\)
b) \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
c) \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}\)
d) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
e) \(\dfrac{\sqrt{a}-\sqrt{b}^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\) với a > 0, b > 0
Bài 1: Rút gọn
a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,
b)\(\sqrt{4+\sqrt{15}}\)+\(\sqrt{4-\sqrt{15}}\)-\(2\sqrt{3-\sqrt{5}}\)
c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
f)D=\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
Mình sửa lại để m.n dễ nhìn hơn!
rút gọn :
a)\(\left(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}+\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
b) \(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}+\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
c) \(\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\dfrac{6}{2-\sqrt{10}}+\sqrt{67+12\sqrt{7}}\)
d) \(\left(\dfrac{\sqrt{5}}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{6}{2-\sqrt{2}}\right).\sqrt{17-12\sqrt{2}}\)
Tính giá trị các biểu thức:
a.\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\sqrt{3}\)
b.\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
c.\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
d.\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
Rút gọn:
a) \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}\right)}{\sqrt{30}}\)
b) \(\dfrac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
c) \(\dfrac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
d) \(\dfrac{10\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3}\left(\sqrt{6}-4\right)}\)
Bài 1 : Rút gọn
a) \(\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}\) b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\) c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) d) \(\sqrt{9+\sqrt{17}}\). \(\sqrt{9-\sqrt{17}}\)
Rút gọn:
a) \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}\right)}{\sqrt{30}}\)
b) \(\dfrac{5\sqrt{7}-4\sqrt{35}+7\sqrt{5}}{\sqrt{35}}\)
c) \(\dfrac{6\sqrt{6}-2\sqrt{12}+3-\sqrt{2}}{2\sqrt{6}+1}\)
d) \(\dfrac{10\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3\left(\sqrt{6}-4\right)}}\)
Rút gọn \(A=\left(\sqrt{6+\sqrt{20}}-2\sqrt{3-\sqrt{5}}+\sqrt{15-10\sqrt{2}}\right):\left(2+\sqrt{8}\right)\)
