G
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
9 tháng 6 2019
a) \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(A^2=\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)^2\)
\(A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(A^2=6+2\sqrt{3^2-5}\)
\(A^2=6+4\)
\(A^2=10\)
\(\Rightarrow\orbr{\begin{cases}A=10\\A=-10\end{cases}}\)
Mà \(A>0\Rightarrow A=10\)
b) \(B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(B^2=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
\(B^2=4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
\(B^2=8-2\sqrt{4^2-7}\)
\(B^2=8-6\)
\(B^2=2\)
\(\Rightarrow\orbr{\begin{cases}B=2\\B=-2\end{cases}}\)
Mà \(B< 0\Rightarrow B=-2\)
9 tháng 6 2019
Cách khác :
b) \(4-\sqrt{7}=\frac{8-2\sqrt{7}}{2}=\frac{7-2\sqrt{7}+1}{2}=\left(\frac{\sqrt{7}-1}{\sqrt{2}}\right)^2\)
\(4+\sqrt{7}=\frac{8+2\sqrt{7}}{2}=\frac{7+2\sqrt{7}+1}{2}=\left(\frac{\sqrt{7}+1}{\sqrt{2}}\right)^2\)
do đó : \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{\left(\frac{\sqrt{7}-1}{\sqrt{2}}\right)^2}-\sqrt{\left(\frac{\sqrt{7}+1}{\sqrt{2}}\right)^2}=\frac{\sqrt{7}-1}{\sqrt{2}}-\frac{\sqrt{7}+1}{\sqrt{2}}=-\sqrt{2}\)
tương tự câu a.
12 tháng 12 2019
a) \(\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=\sqrt{1}=1\)
12 tháng 12 2019
b)
Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=8-6=2\)
\(\Rightarrow B=\sqrt{2}\)
22 tháng 7 2016
a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
NN
28 tháng 10 2020
a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\)
\(=\frac{\sqrt{2\left(4-\sqrt{7}\right)}-\sqrt{2\left(4+\sqrt{7}\right)}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}+2}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+2}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|+2}{\sqrt{2}}=\frac{\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)+2}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1+2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)
b) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\frac{\sqrt{2\left(6+\sqrt{11}\right)}-\sqrt{2\left(6-\sqrt{11}\right)}+3.2}{\sqrt{2}}\)
\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}+6}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}+6}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-1\right|+6}{\sqrt{2}}\)
\(=\frac{\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)+6}{\sqrt{2}}\)
\(=\frac{\sqrt{11}+1-\sqrt{11}+1+6}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}\)
CM
31 tháng 7 2019
mình nghĩ bài này sai đề,
ĐÚng phải là\(\sqrt[3]{2+\sqrt{3}}\)
( KHÔNG CHẮC NỮA :D )
3 tháng 9 2019
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(\sqrt{3}+2\right)}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
= \(\sqrt{4+\sqrt{5\left(\sqrt{3}+5-\sqrt{3}\right)}}\)
= \(\sqrt{4+\sqrt{25}}\)
= \(\sqrt{4+5}=3\)
28 tháng 6 2019
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
25 tháng 6 2018
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}.\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3+4+2\sqrt{12}}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{25+3-2.\sqrt{25.3}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{\left(\sqrt{25}-\sqrt{3}\right)^2}}\)
\(\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(\sqrt{25}=5\)
Ép thành hằng đẳng thức trong căn
Gợi ý.
\(7+4\sqrt{3}=2^2+2.2.\sqrt{3}+3=\left(2+\sqrt{3}\right)^2\)
\(7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)
Tự làm.