so sanh\(\sqrt{24}\)+\(\sqrt{35}\)va 11
so sanh\(\sqrt{24+\sqrt{35}và11}\)
\(\sqrt{24+\sqrt{35}}< \sqrt{25+\sqrt{36}}=\sqrt{5+6}=\sqrt{11}< 11\)
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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\(\sqrt{24}+\sqrt{63}+3\)va 16 .so sanh
Ta có \(16=5+8+3=\sqrt{25}+\sqrt{64}+3.\)
do : \(25>24\Rightarrow\sqrt{25}>\sqrt{24}\); \(64>63\Rightarrow\sqrt{64}>\sqrt{63}\)
=> \(\sqrt{25}+\sqrt{64}+3>\sqrt{24}+\sqrt{63}+3\)
=> \(\sqrt{24}+\sqrt{63}+3< 16\)
ta có căn64>căn63 (1)
căn25>căn24 (2)
167>3 (3)
cộng vế theo vế (1);(2);(3)
=>căn64+căn25+167=16>căn24+căn63+3
1`)So Sanh
a)\(\sqrt{24}+\sqrt{45}\) va 12
b)\(\sqrt{37}-\sqrt{15}\)va 2
giup mk voi nhe
a,Ta có:
\(\left(\sqrt{24}+\sqrt{45}\right)^2=24+45=69\)
\(12^2=144\)
Do 69<144 nên ...
b,tương tự ý a
a ) Ta co \(\sqrt{24}+\sqrt{45}< \sqrt{25}+\sqrt{49}=5+7=12\)
vay \(\sqrt{24}+\sqrt{45}< 12\)
b)ta co \(\sqrt{37}-\sqrt{15}>\sqrt{4}-\sqrt{0}=2-0=2\)
vay \(\sqrt{37}-\sqrt{15}>2\)
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 \(\sqrt{11}+2\ge3+\sqrt{3}\)
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 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 : \(\sqrt{2016}-\sqrt{2015}va\sqrt{\sqrt{2015}-}\sqrt{2014}\)
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 \(\sqrt{11}-\sqrt{7}< \sqrt{7}-\sqrt{3}\)