a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
giúp mk tính
a,\(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)
b,(\(\sqrt{5}+\sqrt{2}\)) (\(3\sqrt{2}-1\))
c,\(3\sqrt{50}-2\sqrt{75}-4\dfrac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\dfrac{1}{3}}\)
d, \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)
e, \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)
f, \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}-\dfrac{20}{\sqrt{10}}\)
bài 2
a, \(\sqrt{9-4\sqrt{5}}\)
b,\(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
c\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
d, \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
e,\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)+\(\dfrac{\sqrt{3}+\sqrt{5}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
f, \(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
1)tính
a)\(\left(\dfrac{1}{5}\sqrt{500}-3\sqrt{45}+5\sqrt{20}\right):\sqrt{5}\)
b)\(\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right).\sqrt{\dfrac{1}{48}}\)
c)\(\left(\dfrac{2\sqrt{3}+3}{\sqrt{3}+2}+\dfrac{2\sqrt{2}}{\sqrt{2}+1}\right):\left(\sqrt{12}+\sqrt{18}\right)\)
Trục căn thức ở mẫu
a)\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}-\sqrt{7}}\)
b)\(\dfrac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
c)\(\dfrac{3+4\sqrt{3}}{\left(\sqrt{6}+\sqrt{2}\right)-\sqrt{5}}\)
Chứng minh các đẳng thức:
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)=1
b)\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)-1 =0
c) \(\sqrt{26+15\sqrt{3}}+\sqrt{26-15\sqrt{3}}-5\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}\)
Rút gọn các biểu thức sau:
a) \(\dfrac{3-2\sqrt{2}}{1-\sqrt{2}}\) b)\(\dfrac{5\sqrt{6}-15}{6-2\sqrt{6}}\)
c) \(\sqrt{\left(3-2\sqrt{2}\right)\left(4-2\sqrt{3}\right)}\)
d) \(\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}\)
I. Thực hiện phép tính
a. \(\left(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}-\dfrac{2-\sqrt{5}}{2+\sqrt{5}}\right)\cdot\dfrac{\sqrt{2}}{23}\)
b. \(\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}\)
Cho a,b,c>0 tm a+b+c=5. \(\sqrt{a}+\sqrt{b}+\sqrt{c}=3\).
C/m\(\dfrac{\sqrt{a}}{2+a}+\dfrac{\sqrt{b}}{2+b}+\dfrac{\sqrt{c}}{2+c}=\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
rút gọn biểu thức sau:
a.\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
b.\(A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\) với a\(\ge\)0; a\(\ne25\)
cho a,b,c dương thỏa mãn \(a+b+c=5\) và \(\sqrt{a}+\sqrt{b}+\sqrt{c}=3\). CMR: \(\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
