a: \(1-\sqrt{3}=\dfrac{-2}{1+\sqrt{3}}\)
\(2-\sqrt{6}=\dfrac{-2}{2+\sqrt{6}}\)
mà 1+căn 3<2+căn 6
nên 1-căn 3>2-căn 6
b: \(\left(2-\sqrt{2}\right)^2=6-4\sqrt{2}\)
(1/2)^2=1/4
mà 6-4căn 2-1/4>0
nên 2-căn 2>1/2
a: \(1-\sqrt{3}=\dfrac{-2}{1+\sqrt{3}}\)
\(2-\sqrt{6}=\dfrac{-2}{2+\sqrt{6}}\)
mà 1+căn 3<2+căn 6
nên 1-căn 3>2-căn 6
b: \(\left(2-\sqrt{2}\right)^2=6-4\sqrt{2}\)
(1/2)^2=1/4
mà 6-4căn 2-1/4>0
nên 2-căn 2>1/2
So sánh:
a) \(1-\sqrt{3}\) và \(2-\sqrt{6}\)
b) \(2-\sqrt{2}\) và \(\dfrac{1}{2}\)
c) \(\sqrt{2003}+\sqrt{2005}\) và \(2\sqrt{2004}\)
bài 45:so sánh
a)3\(\sqrt{3}\) và \(\sqrt{12}\)
b)7 và 3\(\sqrt{5}\)
c)\(\dfrac{1}{3}\sqrt{51}\) và \(\dfrac{1}{5}\sqrt{150}\)
d)\(\dfrac{1}{2}\sqrt{6}\) và \(6\sqrt{\dfrac{1}{2}}\)
So sánh:
a. \(3\sqrt{3}\) và \(\sqrt{12}\) ; b. 7 và \(3\sqrt{5};\)
c. \(\dfrac{1}{3}\sqrt{51}\) và \(\dfrac{1}{5}\sqrt{150};\) d. \(\dfrac{1}{2}\sqrt{6}\) và \(6\sqrt{\dfrac{1}{2}}.\)
Bài 1 : Tính :
a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\)
c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)
d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\)
f) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}-\sqrt{3}}\right):\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
Bài 2 : Rút gọn :
a) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
c) \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
so sánh
\(\sqrt{2003}+\sqrt{2005}\) và \(2\sqrt{2004}\)
bài 1 rút gọn
a) √98 - √72 + 0,5√8
b) √9a - √16a +√49
bài 2 so sánh
a) 2√7 và 3√2
b) 5 và 2 + √2
bài 3 khử mẫu
a)\(\sqrt{\dfrac{2}{3}}\)
b)\(\dfrac{x}{y}\). \(\sqrt{\dfrac{y}{x}}\)
a) \(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}}\)
b) \(\dfrac{3}{\sqrt{2}-1}+\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{3}+1}\)
c) \(\dfrac{3}{\sqrt{5}-2}-\dfrac{3}{\sqrt{5}+2}\)
Rút gọn:
a)\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}\)
b)\(\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\)
c)\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\div\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d)\(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
e)\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\times\left(\sqrt{6}+11\right)\)
Lm nhanh giúp mk nhé, mk đang cần gấp!
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)