![](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,Ta có : \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)
Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
b, Đặt A = \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)
\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)
Vậy (*) = 0
1:
Ta có: \(\sqrt{2}-\sqrt{6}\)
\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)
\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Ta có 4=2 căn 4
Căn 4<căn 5
=> 2 căn 5 >4
2. Ta có 3^2=9 =16-7=16-căn 49
( căn 15 -1)^2
= 15 -2 căn 15 +1= 16-2 căn 15 =16- căn 60
Căn 60>căn49
=> 3> căn 15 -1
3. Ta có 6^2=36=27+9= 27+ căn 81
(căn 26 +1)^2=26 +2 căn 26 +1=27+ 2 căn 26 =27+ căn 52
Căn 52< căn 81
=> 6> căn 26+1
4. Ta có (căn 2 -2)^2 =2- 4 căn 2+4=6- 4 căn 2
(căn 3 -3 )^2 = 3 -6 căn 3 +9= 12- 6 căn 3
Lại có 8 căn 2 =căn 128
6 căn 3 =căn 108
=> (căn 3 -3)^2> 2(căn 2 -2)^2
=> căn 3 -3 > căn 2-2
\(2\sqrt{5}>4\)
\(3< \sqrt{15-1}\)
\(6>\sqrt{26-1}\)
\(\sqrt{2-2}=\sqrt{3-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-3\sqrt{3}=-\sqrt{27}\)
\(-2\sqrt{7}=-\sqrt{28}\)
mà 27<28
nên \(-3\sqrt{3}>-2\sqrt{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)
Ta có: \(2-\sqrt{3}=\frac{1}{2+\sqrt{3}}\)
\(7-4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)
<=> \(\frac{1}{\left(2+\sqrt{3}\right)^x}+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)
<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)
<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^{2x}=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)
Đặt: \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=t\)
Ta có pt ẩn t: \(1+t^2=4t\)
<=> \(t^2-4t+1=0\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{3}\\t=2+\sqrt{3}\end{cases}}\)
+) Với \(t=2+\sqrt{3}\), ta có:
\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=2+\sqrt{3}\)
<=> \(\left(2+\sqrt{3}\right)^x=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(2+\sqrt{3}\right)^2\)
<=> x=2
Trường hợp còn lại em làm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2\sqrt{2}+\sqrt{3}\)
\(B=\dfrac{2\sqrt{2}}{1+\sqrt{2-\sqrt{3}}}=\dfrac{4}{2+\sqrt{3}-1}=\dfrac{4}{\sqrt{3}+1}=2\sqrt{3}-2\)
=>A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2>\sqrt{3}\)
trả lời
>
chúc bn
hc tốt