Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có : \(\left(\sqrt{11}+\sqrt{13}\right)^2=11+2\sqrt{11.13}+13=24+2\sqrt{143}\)
\(\left(2.\sqrt{12}\right)^2=4.12=24+2.\sqrt{144}\)
mà \(\sqrt{144}>\sqrt{143}\Rightarrow24+2\sqrt{144}>24+2\sqrt{143}\Rightarrow\left(2.\sqrt{12}\right)^2>\left(\sqrt{11}+\sqrt{13}\right)^2\)
\(2.\sqrt{12}>\sqrt{11}+\sqrt{13}\)
b) Ta có : \(\left(\sqrt{69}-\sqrt{68}\right)-\left(\sqrt{68}-\sqrt{69}\right)\)
\(\Leftrightarrow\sqrt{69}+\sqrt{67}-2\sqrt{68}\)
Từ kq câu a \(\Rightarrow\sqrt{69}+\sqrt{67}< 2\sqrt{68}\)
\(\Rightarrow\sqrt{69}+\sqrt{67}-2\sqrt{68}< 0\)
\(\Rightarrow\left(\sqrt{69}-\sqrt{68}\right)-\left(\sqrt{68}-\sqrt{67}\right)< 0\)
\(\Rightarrow\sqrt{69}-\sqrt{68}< \sqrt{68}-\sqrt{67}\)
a) \(\left(\sqrt{11}+\sqrt{14}\right)^2=25+\sqrt{154}\)
\(\left(2\sqrt{12}\right)^2=24+\sqrt{144}\)
Vậy \(2\sqrt{12}< \sqrt{11}+\sqrt{14}\)
b) \(\left(\sqrt{a+1}+\sqrt{a+3}\right)^2=2a+4+\sqrt{\left(a+1\right)\left(a+3\right)}\)
\(\left(2\sqrt{a+2}\right)^2=2a+4+\sqrt{\left(a+2\right)\left(a+2\right)}\)
Vậy \(\sqrt{a+1}+\sqrt{a+3}< 2\sqrt{a+2}\)
a) Ta có : \(5>2\Rightarrow\sqrt{5}>\sqrt{2}\)
b) Vì \(8>5\Rightarrow\sqrt{8}>\sqrt{5}\Rightarrow2\sqrt{2}>5\)
c) VÌ \(-32>-45\Rightarrow-\sqrt{32}>-\sqrt{45}\Rightarrow-4\sqrt{2}>-\sqrt{5}\)
d) Vì \(12< 18\Rightarrow\sqrt{12}< \sqrt{18}\Leftrightarrow2\sqrt{3}< 3\sqrt{2}\)
\(a\)
\(\sqrt{7}+\sqrt{15}\)
\(=\sqrt{7+15}\)
\(=4,69\)
\(4,69< 7\)
\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)
\(b\)
\(\sqrt{7}+\sqrt{15}+1\)
\(=\sqrt{7+15}+1\)
\(=4,69+1\)
\(=5,69\)
\(\sqrt{45}\)
\(=6,7\)
\(5,69< 6,7\)
\(\Rightarrow\)\(\sqrt{7}+\sqrt{15}+1\)\(< \)\(\sqrt{45}\)
\(c\)
\(\frac{23-2\sqrt{19}}{3}\)
\(=\frac{22.4,53}{3}\)
\(=\frac{95,7}{3}\)
\(=31,9\)
\(\sqrt{27}\)
\(=5,19\)
\(31,9>5,19\)
\(\text{}\Rightarrow\text{}\text{}\)\(\frac{23-2\sqrt{19}}{3}\)\(>\sqrt{27}\)
\(d\)
\(\sqrt{3\sqrt{2}}\)
\(=\sqrt{3.1,41}\)
\(=\sqrt{4,23}\)
\(=2,05\)
\(\sqrt{2\sqrt{3}}\)
\(=\sqrt{2.1,73}\)
\(=\sqrt{3,46}\)
\(=1,86\)
\(2,05>1,86\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
\(Học \) \(Tốt !!!\)
a) Ta có : \(\sqrt{7}< \sqrt{9}=3;\sqrt{15}< \sqrt{16}=4\)
Do đó : \(\sqrt{7}+\sqrt{15}< 3+4=7\)
b) Ta có : \(\sqrt{17}>\sqrt{16}=4;\sqrt{5}>\sqrt{4}=2\)
\(\Rightarrow\sqrt{17}+\sqrt{5}+1>4+2+1=7\)
Lại có : \(\sqrt{45}< \sqrt{49}< 7\)
Do đó : \(\sqrt{17}+\sqrt{5}+1>\sqrt{45}\)
c) Ta thấy : \(\sqrt{19}>\sqrt{16}=4\)
\(\Rightarrow2\sqrt{19}>2.4=8\)
\(\Rightarrow-2\sqrt{19}< -8\)
\(\Rightarrow23-2\sqrt{19}< 23-8=15\)
\(\Rightarrow\frac{23-2\sqrt{19}}{3}< 5\). Mặt khác : \(\sqrt{27}>\sqrt{25}=5\)
Nên : \(\frac{23-2\sqrt{19}}{3}< \sqrt{27}\)
d) Vì : \(18>12>0\Rightarrow\sqrt{18}>\sqrt{12}>0\)
\(\Leftrightarrow3\sqrt{2}>2\sqrt{3}>0\)
\(\Rightarrow\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
cả hai bài đều giải bằng cách bình phương cả hai vế rồi so sánh
So sánh từng vế:
\(\sqrt{15}+1=4,872983346\)
\(\sqrt{24}=4,898979486\)
Vậy: \(\sqrt{15}+1< \sqrt{24}\)
\(\sqrt{2002}+\sqrt{2004}=89,50977321\)
\(2\sqrt{2005}=89,5545271\)
Vậy \(\sqrt{2002}+\sqrt{2004}< 2\sqrt{2005}\)
P/s: Ko chắc
Lời giải:
\(\sqrt{20152015}+\sqrt{20152017}-2\sqrt{20152016}=(\sqrt{20152015}-\sqrt{20152016})+(\sqrt{20152017}-\sqrt{20152016})\)
\(=\frac{-1}{\sqrt{20152015}+\sqrt{20152016}}+\frac{1}{\sqrt{20152017}+\sqrt{20152016}}\)
Dễ thấy: $0< \sqrt{20152015}+\sqrt{20152016}<\sqrt{20152017}+\sqrt{20152016}}$
$\Rightarrow \frac{1}{\sqrt{20152015}+\sqrt{20152016}}>\frac{1}{\sqrt{20152017}+\sqrt{20152016}}$
$\Rightarrow \frac{-1}{\sqrt{20152015}+\sqrt{20152016}}+\frac{1}{\sqrt{20152017}+\sqrt{20152016}}< 0$
$\Rightarrow \sqrt{20152015}+\sqrt{20152017}< 2\sqrt{20152016}$
Lời giải:
Ta có:
$\sqrt{2015.2015}+\sqrt{2015.2017}=\sqrt{2015}(\sqrt{2015}+\sqrt{2017})$
Mà:
$(\sqrt{2015}+\sqrt{2017})^2=4032+2\sqrt{2015.2017}$
$=4032+2\sqrt{(2016-1)(2016+1)}=4032+2\sqrt{2016^2-1}$
$< 4032+2\sqrt{2016^2}=4.2016$
$\Rightarrow \sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}$
$\Rightarrow \sqrt{2015.2015}+\sqrt{2015.2017}=\sqrt{2015}(\sqrt{2015}+\sqrt{2017})< \sqrt{2015}.2\sqrt{2016}$
Vậy......
không phải nhân đâu bạn ơi. đó là 1 số có 6 chữ số đấy
Kanzaki Mizuki: ok mình sửa lại bài rồi đó bạn.