![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\\ \Leftrightarrow6+2\sqrt{2}< 3+6=9\\ b,\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}\\ 2^2=4=14-10\\ \left(2\sqrt{33}\right)^2=132>100=10^2\Leftrightarrow-2\sqrt{33}< -10\\ \Leftrightarrow\sqrt{11}-\sqrt{3}< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) \(-x^2+4x-2\)
\(=-\left(x^2-4x+2\right)\)
\(=-\left(x^2-4x+4-2\right)\)
\(=-\left(x-2\right)^2+2\)
Ta có: \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)
Vậy: GTLN của bt là 2 tại x=2
b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))
Mà: \(\sqrt{2x^2-3}\ge0\forall x\)
Dấu "=" xảy ra:
\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)
Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)
...
1:
b: \(4\sqrt{5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{75}\)
=>\(4\sqrt{5}>5\sqrt{3}\)
=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)
c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)
=>\(3-2\sqrt{5}< 1-\sqrt{5}\)
d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)
\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)
=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)
=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)
e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Có:
\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)
Vì \(\sqrt{117}>\sqrt{116}\) nên \(3\sqrt{13}>2\sqrt{29}\)
b)
Có:
\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)
\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)
Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\) nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)
c)
Có:
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)
Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)
d)
Có:
\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)
\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)
lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)
\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2=\sqrt{4}>\sqrt{3}\)
b) \(6=\sqrt{36}< \sqrt{41}\)
c) \(7=\sqrt{49}>\sqrt{47}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)
b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)
a: \(2\sqrt{5}=\sqrt{20}\)
\(5=\sqrt{25}\)
mà 20<25
nên \(2\sqrt{5}< 5\)
b: \(\dfrac{1}{3}\cdot\sqrt{16}=\sqrt{\dfrac{1}{9}\cdot16}=\sqrt{\dfrac{16}{9}}\)
\(\sqrt{12}=\sqrt{\dfrac{108}{9}}\)
mà 16<9
nên \(\dfrac{1}{3}\sqrt{16}< \sqrt{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$
$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.
$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$
$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$
$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(2\sqrt{3}=\sqrt{4\cdot3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{9\cdot2}=\sqrt{18}\)
mà \(\sqrt{12}< \sqrt{18}\)(vì 12<18)
nên \(2\sqrt{3}< 3\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{3}+1\right)^2=8+4\sqrt{3}+1=9+4\sqrt{3}\)
\(4^2=16=9+7\)
mà \(4\sqrt{3}< 7\left(\sqrt{48}< \sqrt{49}\right)\)
nên \(\left(2\sqrt{3}+1\right)^2< 4^2\)
hay \(2\sqrt{3}+1< 4\)
c) Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)
\(\sqrt{2014}-\sqrt{2013}=\dfrac{1}{\sqrt{2014}+\sqrt{2013}}\)
Ta có: \(\sqrt{2015}+\sqrt{2014}>\sqrt{2013}+\sqrt{2014}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2015}+\sqrt{2014}}< \dfrac{1}{\sqrt{2013}+\sqrt{2014}}\)
hay \(\sqrt{2015}-\sqrt{2014}< \sqrt{2014}-\sqrt{2013}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c.
(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})
Mà:
\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)
\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)
\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)
Lời giải:
a.
$5+\sqrt{2}>5+\sqrt{1}=6$
$4+\sqrt{3}< 4+\sqrt{4}=6$
$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$
b.
$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$
Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$
a, Ta có: \(\left(\sqrt{2}+\sqrt{3}\right)^2\)= \(2+2\sqrt{6}+3=5+2\sqrt{6}\)
Lại có \(3^2=9=5+4\)mà \(2\sqrt{6}>4\)
suy ra \(\left(\sqrt{2}+\sqrt{3}\right)^2>9\)
suy ra \(\sqrt{2}+\sqrt{3}>3\)
b, Ta có: \(\left(\sqrt{11}-\sqrt{3}\right)^2=11-2\sqrt{33}+3=14-2\sqrt{33}\)
Lại có: \(2^2=4=14-10\)mà \(2\sqrt{33}>10\)
suy ra \(\left(\sqrt{11}-\sqrt{3}\right)^2< 2^2\)
suy ra \(\sqrt{11}-\sqrt{3}< 2\)
#)Giải :
a) √2 +√3 = √( √2 + √3 )2 = √( 5 + 2√6 ) = √( 5 + √24 )
3 = √9 = √( 5 + √16 )
=> √2 + √3 > 3