tính A=\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
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EO
tính x=\(\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)
y=\(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)
2
15 tháng 1 2021
Ta có: \(x=\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)
\(=\sqrt{49-2\cdot7\cdot4\sqrt{3}+48}+\sqrt{48+2\cdot4\sqrt{3}\cdot2+4}\)
\(=\sqrt{\left(7-4\sqrt{3}\right)^2}+\sqrt{\left(4\sqrt{3}+2\right)^2}\)
\(=\left|7-4\sqrt{3}\right|+\left|4\sqrt{3}+2\right|\)
\(=7-4\sqrt{3}+4\sqrt{3}+2\)
\(=9\)
15 tháng 1 2021
Làm luôn phần y :D
y = \(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)
y = \(\sqrt{33+2.10\sqrt{2}}+\sqrt{24-2.8\sqrt{2}}\)
y = \(\sqrt{33+2.5.2\sqrt{2}}+\sqrt{24-2.4.2\sqrt{2}}\)
y = \(\sqrt{25+2.5.\sqrt{8}+8}+\sqrt{16-2.4.\sqrt{8}+8}\)
y = \(\sqrt{\left(5+\sqrt{8}\right)^2}+\sqrt{\left(4-\sqrt{8}\right)^2}\)
y = |5 + \(\sqrt{8}\)| + |4 - \(\sqrt{8}\)|
y = 5 + \(\sqrt{8}\) + 4 - \(\sqrt{8}\) (Vì 4 > \(\sqrt{8}\) nên 4 - \(\sqrt{8}\) > 0)
y = 9
Vậy y = 9
Chúc bn học tốt!
HN
0
NT
1
1 tháng 12 2016
\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}=\sqrt{8\left(3+2\sqrt{2}\right)}-\sqrt{8\left(3-2\sqrt{2}\right)}\)
\(=\sqrt{8}.\left[\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\right]=\sqrt{8}.\left(\sqrt{2}+1-\sqrt{2}+1\right)=2\sqrt{8}=4\sqrt{2}\)
11 tháng 6 2018
\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
\(=\sqrt{\left(4+2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=4+2\sqrt{2}-4+2\sqrt{2}\)
\(=4\sqrt{2}\)
HH
11 tháng 6 2018
Giải:
\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
\(=\sqrt{8+2.4.2\sqrt{2}+16}-\sqrt{16-2.4.2\sqrt{2}+8}\)
\(=\sqrt{\left(2\sqrt{2}+4\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}+4-\left(4-2\sqrt{2}\right)\)
\(=2\sqrt{2}+4-4+2\sqrt{2}\)
\(=4\sqrt{2}\)
Vậy ...
LT
2
HR
30 tháng 11 2017
\(8\sqrt{2}\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)\)
\(=8\sqrt{2}\left(\sqrt{16+2.4.\sqrt{8}+8}-\sqrt{16-2.4\sqrt{8}+8}\right)\)
\(=8\sqrt{2}\left(\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}\right)\)
\(=8\sqrt{2}\left(4+\sqrt{8}-4+\sqrt{8}\right)\)
\(=8\sqrt{2}.2\sqrt{8}\)
= 64
DT
11 tháng 12 2017
ta có\(8\sqrt{2}\cdot\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)=8\sqrt{2}\cdot\left(\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}\right)=8\sqrt{2}\cdot\left(4+\sqrt{8}-4+\sqrt{8}\right)=8\sqrt{2}\cdot2\sqrt{8}=64\)vây..................
14 tháng 7 2017
A=\(\sqrt{\left(4+\sqrt{8}\right)^2}\)\(-\sqrt{\left(4-\sqrt{8}\right)^2}\)=\(4+\sqrt{8}\)\(-\left(4-\sqrt{8}\right)\)=\(2\sqrt{8}\)
Giờ mình chỉ giải đc câu a thôi để hồi nao mình rảnh giải típ cho
3 tháng 7 2018
\(\sqrt{24-16\sqrt{2}}+\sqrt{12-8\sqrt{2}}=\dfrac{\sqrt{32-2.4.4\sqrt{2}+16}+\sqrt{12-2.4.2\sqrt{2}+16}}{\sqrt{2}}=\dfrac{4\sqrt{2}-4+4-2\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}}=1\)
9 tháng 10 2021
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
\(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}=\left|4+\sqrt{8}\right|-\left|4-\sqrt{8}\right|=4+\sqrt{8}-4+\sqrt{8}=4\sqrt{2}\)
\(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)
\(=\sqrt{8+2.4.2\sqrt{2}+16}-\sqrt{16-2.4.2\sqrt{2}+8}\)
\(=\sqrt{\left(2\sqrt{2}+4\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}+4-4+2\sqrt{2}\)
\(=4\sqrt{2}\)