So sanh:
a, \(2-2\sqrt{3}\) va \(4-\sqrt{15}\)
b, \(\sqrt{11}+2\) va \(3+\sqrt{3}\)
so sanh x va y biet
a) x=\(2\sqrt{7}\)va y=\(3\sqrt{3}\)
b) x=\(6\sqrt{2}\)va y=\(5\sqrt{3}\)
c) x=\(\sqrt{31}-\sqrt{33}\) va y=\(6-\sqrt{11}\)
so sanh : A=\(\sqrt{11+\sqrt{96}}\) va B=\(\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)
\(A=\sqrt{11+\sqrt{96}}>B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)
Đã làm: https://olm.vn/hoi-dap/detail/223607632837.html
So sanh:
a, \(\sqrt{\dfrac{35}{34}}\) va \(\sqrt{\dfrac{71}{70}}\)
b, \(4\sqrt{5}-3\sqrt{2}\) va 5
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so sanh ko dung may tinh
1 )\(\sqrt{3}\) +\(\sqrt{7}\) va 2+ \(\sqrt{6}\)
2) \(\sqrt{7}\) - \(\sqrt{5}\) va \(\sqrt{6}-2\)
3) \(\sqrt{11}-\sqrt{7}vs\sqrt{7}-\sqrt{3}\)
1: \(\left(\sqrt{3}+\sqrt{7}\right)^2=10+2\sqrt{21}\)
\(\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)
mà 2 căn 21<4 căn 6
nên căn 3+căn 7<2+căn 6
2: \(\sqrt{7}-\sqrt{5}=\dfrac{2}{\sqrt{7}+\sqrt{5}}\)
\(\sqrt{6}-2=\dfrac{2}{\sqrt{6}+2}\)
mà \(\sqrt{7}+\sqrt{5}>\sqrt{6}+2\)
nên \(\sqrt{7}-\sqrt{5}< \sqrt{6}-2\)
3: \(\sqrt{11}-\sqrt{7}=\dfrac{4}{\sqrt{11}+\sqrt{7}}\)
\(\sqrt{7}-\sqrt{3}=\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
mà căn 11>căn 3
nên \(\sqrt{11}-\sqrt{7}< \sqrt{7}-\sqrt{3}\)
So sanh :
A=\(\sqrt[3]{2017}+\sqrt[3]{2019}\) va B=\(2\sqrt[3]{2018}\)
bai 4 so sanh cac so thuc
\(\frac{4}{9}va\)0,4(5)
\(\sqrt[2]{3}va\sqrt[3]{2}\)
\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
Xếp theo thứ tự tăng dần: \(21,2\sqrt{7},15\sqrt{3},-\sqrt{123}\) ; \(28\sqrt{2},\sqrt{14},2\sqrt{147},36\sqrt{4}\)
giảm dần: \(6\sqrt{\frac{1}{4}},4\sqrt{\frac{1}{2}},-\sqrt{132},2\sqrt{3},\sqrt{\frac{15}{5}}\); \(-27,4\sqrt{3},16\sqrt{5},21\sqrt{2}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
so sanh
a) 2^90 va 5^36
b) 99^200 va 9999^100
c) 2^150 va 3^100
d) \(\sqrt{26+2}\) va \(\sqrt{26}\)+ \(\sqrt{2}\)
So sánh:
1,\(2-\sqrt{2}va\frac{1}{2}\)
2, \(2\sqrt{3}-5va\sqrt{3}-4\)
3, \(\sqrt{3}-3\sqrt{2}va-4\sqrt{3}+5\sqrt{2}\)
4,\(1-\sqrt{3}va\sqrt{2}-\sqrt{6}\)
5, \(\sqrt{4\sqrt{5}}va\sqrt{5\sqrt{3}}\)
6, \(\sqrt{\sqrt{6}-\sqrt{5}}-\sqrt{\sqrt{3}-\sqrt{2}}va..0\)
7, \(-2\sqrt{\frac{1}{2}\sqrt{5}}va-3\sqrt{\frac{1}{3}\sqrt{2}}\)