so sanh 3+\(2\sqrt{2}\)va 7-\(\sqrt{3}\)
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 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, \(2-2\sqrt{3}\) va \(4-\sqrt{15}\)
b, \(\sqrt{11}+2\) va \(3+\sqrt{3}\)
a) \(2-2\sqrt{3}\) và \(4-\sqrt{15}\)
Giả sử : \(2-2\sqrt{3}\ge4-\sqrt{15}\)
⇔ \(\sqrt{15}-2\sqrt{3}\ge2\)
⇔ \(\left(\sqrt{15}-2\sqrt{3}\right)^2\ge2^2\)
⇔ 15 - \(12\sqrt{5}+12\) ≥ 4
⇔ 27 -4 ≥ \(12\sqrt{5}\)
⇔ 23 ≥ \(12\sqrt{5}\)
⇔ \(23^2\) ≥ \(\left(12\sqrt{5}\right)^2\)
⇔ 529 ≥ 720 (sai)
Vậy 2 - \(2\sqrt{3}< 4-\sqrt{15}\)
b) \(\sqrt{11}+2\) và \(3+\sqrt{3}\)
Giả sử : \(\sqrt{11}+2\le3+\sqrt{3}\)
⇔ \(\sqrt{11}-\sqrt{3}\le1\)
⇔ \(\left(\sqrt{11}-\sqrt{3}\right)^2\le1\)
⇔ 14 - \(2\sqrt{33}\) ≤ 1
⇔ 13 ≤ \(2\sqrt{33}\)
⇔ \(13^2\le\left(2\sqrt{33}\right)^2\)
⇔ 169 ≤ 132 (sai)
Vậy \(\sqrt{11}+2\ge3+\sqrt{3}\)
1) So sanh
\(\frac{7}{\sqrt{1}}+\frac{7}{\sqrt{2}}+\frac{7}{\sqrt{3}}+...+\frac{7}{\sqrt{90}}+\frac{7}{\sqrt{100}}.\)va 70
2)Tinh
\(\frac{1}{x^2+3x+2}+\frac{2x}{x^3+4x^2+4x}+\frac{1}{x^2+5x+6}\)
bai 4 so sanh cac so thuc
\(\frac{4}{9}va\)0,4(5)
\(\sqrt[2]{3}va\sqrt[3]{2}\)
So sanh :
A=\(\sqrt[3]{2017}+\sqrt[3]{2019}\) va B=\(2\sqrt[3]{2018}\)
so sanh
\(2\sqrt{5}-5\)va \(\sqrt{5}-3\)
\(\sqrt{17}+\sqrt{26}\) va 9
a ) \(2\sqrt{5}-5\) và \(\sqrt{5}-3\)
Ta có ; \(2\sqrt{5}-5-\left(\sqrt{5}-3\right)\)
\(=\sqrt{5}-8\)
\(=\sqrt{5}-\sqrt{64}< 0\)
\(\Rightarrow2\sqrt{5}-5< \sqrt{5}-3\)
Vậy .................
b ) \(\sqrt{17}+\sqrt{26}\) và 9
Ta có :
\(\sqrt{17}>\sqrt{16}\)
\(\sqrt{26}>\sqrt{25}\)
\(\Rightarrow\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}=4+5=9\)
Vậy ...
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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