Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1+\sqrt{3}}{\sqrt{3}-1}=\frac{\left(1+\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=2+\sqrt{3}\)
\(\frac{2}{\sqrt{2}-1}=\frac{2\sqrt{2}+2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}+2=\sqrt{8}+2\)
\(\Rightarrow\frac{2}{\sqrt{2}-1}>\frac{1+\sqrt{3}}{\sqrt{3}-1}\)
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 \(\Rightarrow\sqrt{69}+\sqrt{67}< 2\sqrt{68}\)
\(\Rightarrow\sqrt{69}+\sqrt{67}-2\sqrt{68}< 0\)
\(\Rightarrow\left(\sqrt{69}-\sqrt{68}\right)-\left(\sqrt{68}-\sqrt{67}\right)< 0\)
\(\Rightarrow\sqrt{69}-\sqrt{68}< \sqrt{68}-\sqrt{67}\)
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
a: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)
\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)
mà \(2\sqrt{22}< 15+10\sqrt{3}\)
nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)
b: \(\left(\sqrt{8}+\sqrt{11}\right)^2=19+2\cdot\sqrt{88}=19+\sqrt{352}\)
\(\left(\sqrt{38}\right)^2=19+19=19+\sqrt{361}\)
mà 352<361
nên \(\sqrt{8}+\sqrt{11}< \sqrt{38}\)
a, \(\left(\sqrt{2\sqrt{2}}\right)^2=2\sqrt{2}< 2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
=> \(2\sqrt{2}< \sqrt{2}+1\)( vì \(2\sqrt{2}>0,\sqrt{2}+1>0\))
b, \(1=\left(\sqrt{12}-\sqrt{11}\right)\left(\sqrt{12}+\sqrt{11}\right)\)
=> \(\sqrt{12}-\sqrt{11}=\frac{1}{\sqrt{12}+\sqrt{11}}\)
Tương tự: \(\sqrt{11}-\sqrt{10}=\frac{1}{\sqrt{11}+\sqrt{10}}\)
Do \(\sqrt{12}+\sqrt{11}>\sqrt{11}+\sqrt{10}\)<=> \(\sqrt{12}-\sqrt{11}=\frac{1}{\sqrt{12}+\sqrt{11}}< \frac{1}{\sqrt{11}+\sqrt{10}}=\sqrt{11}-\sqrt{10}\)
=> \(\sqrt{12}-\sqrt{11}< \sqrt{11}-\sqrt{10}\)