\(sosanh\frac{\sqrt{8}}{3}va\frac{3}{4}\)
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\(sosanh\frac{n}{n+1}va\frac{3n+1}{6n+3}\)
\(sosanh\frac{-\sqrt{10}}{2}va-2\sqrt{5}\)
\(\frac{\sqrt{10}}{2}=\sqrt{\frac{10}{4}}-2\sqrt{5}\)
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\(\frac{-\sqrt{10}}{2}=\frac{-\sqrt{2.5}}{2}=\frac{-\sqrt{2}.\sqrt{5}}{2}=-\frac{\sqrt{5}}{\sqrt{2}}=-\sqrt{\frac{5}{2}}>-2\sqrt{5}\)
đúng k
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\(sosanh\frac{99}{-100}va`\frac{-102}{101}\)
vì : \(\frac{99}{-100}< -1\)và \(\frac{-102}{101}>-1\)
=> \(\frac{99}{-100}>\frac{-102}{101}\)
k nha!
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Ta có:
\(\frac{99}{-100}< -1\) ; \(\frac{-102}{101}>-1\)
\(\Rightarrow\frac{99}{-100}< \frac{-102}{101}\)
nha bn
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so sanh : 99/-100 va -102/101
ta co: -99/100=9999/10100
-102/101=10200/10100
vi 10200>9999 => 10200/10100>9999/10100 => 99/-100 < -102/101
chi co cach nay( quy dong mau) hoac quy dong tu moi co the giai dc.
chuc ban hoc tot! ket ban nha!
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left(5+4sqrt{2}right)left(3+2sqrt{1+sqrt{2}}right)left(3-2sqrt{1+sqrt{2}}right) sqrt{frac{9}{4}-sqrt{2}} Sosanh2sqrt{27}vasqrt{147}
2sqrt{15}vasqrt{59}
2sqrt{2}-1va2
frac{sqrt{3}}{2}va1
-frac{sqrt{10}}{2}va-2sqrt{5}
sqrt{6}-1va3
2sqrt{5}-5sqrt{2}va1
frac{sqrt{8}}{3}vafrac{3}{4}
-2sqrt{6}va-sqrt{23}
2sqrt{6}-2va3
sqrt{111}-7va4
Xếp theo thứ tự tăng dần: 21,2sqrt{7},15sqrt{3},-sqrt{123} ; 28sqrt{2},sqrt{14},2sqrt{147},36sqrt{4}
giảm dần: 6sqrt{frac{1}{4}},4sqrt{frac{1}{2}...
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\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\\ \\ \\ \sqrt{\frac{9}{4}-\sqrt{2}}\\ \\ \\ Sosanh2\sqrt{27}va\sqrt{147}\\ \\ \\ 2\sqrt{15}va\sqrt{59}\\ \\ \\ 2\sqrt{2}-1va2\\ \\ \\ \frac{\sqrt{3}}{2}va1\\ \\ \\ -\frac{\sqrt{10}}{2}va-2\sqrt{5}\\ \\ \\ \sqrt{6}-1va3\\ \\ \\ 2\sqrt{5}-5\sqrt{2}va1\\ \\ \\ \frac{\sqrt{8}}{3}va\frac{3}{4}\\ \\ \\ -2\sqrt{6}va-\sqrt{23}\\ \\ \\ 2\sqrt{6}-2va3\\ \\ \\ \sqrt{111}-7va4\)
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, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
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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: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
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\(\left(1\right)sosanh\frac{n}{n+3}v\text{à}\frac{n-1}{n+4}\)
nhanh ho mik mai cha bai roi
ai nhanh nhat mik k
nhung phai dung
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\(a,b,n\in N\cdot sosanh\frac{a+n}{b+n}va\frac{a}{a}\)
a,Cho a,b,c duong va \(a^2+b^2+c^2\)=3. Tim Min cua P= \(\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)
b,Cho x,y,z>0 va x+y+z=6. C/m \(8^x+8^y+8^z\ge4^{x+1}+4^{y+1}+4^{z+1}\)
a/
-Cauchy-Schwar
\(P=\sum\frac{a^4}{a\sqrt{b^2+3}}\ge\frac{\left(\sum a^2\right)^2}{\sum a\sqrt{b^2+3}}\)
Côsi: \(\sum a\sqrt{b^2+3}=\frac{1}{2}\sum2a.\sqrt{b^2+3}\le\frac{1}{2}.\sum\frac{\left(2a\right)^2+b^2+3}{2}=\frac{1}{4}.\left[5\left(a^2+b^2+c^2\right)+3.3\right]=6\)
\(\Rightarrow P\ge\frac{3^2}{6}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1.
b/
Côsi: \(8^x+8^x+64\ge3\sqrt[3]{8^x.8^x.64}=12.4^x\Rightarrow8^x\ge6.4^x-32\)
\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-96\)
\(4^x+4^y+4^z\ge3\sqrt[3]{4^{x+y+z}}=3\sqrt[3]{4^6}=48\)
\(\Rightarrow-2\left(4^x+4^y+4^z\right)\le-96\)
\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-2\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)
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truc can thuc va tinh
a) \(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\)
b) \(\frac{4}{\sqrt{5}-\sqrt{2}}+\frac{3}{\sqrt{5}-2}-\frac{2}{\sqrt{3}-2}+\frac{\sqrt{3}-1}{6}\)
Bạn xem hộ mk đề cậu b nhé căn 5- căn 2 hay là căn 5 - 2
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So sánh:1,2-sqrt{2}vafrac{1}{2}2, 2sqrt{3}-5vasqrt{3}-43, sqrt{3}-3sqrt{2}va-4sqrt{3}+5sqrt{2}4,1-sqrt{3}vasqrt{2}-sqrt{6}5, sqrt{4sqrt{5}}vasqrt{5sqrt{3}}6, sqrt{sqrt{6}-sqrt{5}}-sqrt{sqrt{3}-sqrt{2}}va..07, -2sqrt{frac{1}{2}sqrt{5}}va-3sqrt{frac{1}{3}sqrt{2}}
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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, \(-2\sqrt{\frac{1}{2}\sqrt{5}}va-3\sqrt{\frac{1}{3}\sqrt{2}}\)