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30 tháng 10 2018
\(B=\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}-\sqrt{5}-\sqrt{7}}{5-7}=\frac{-2\sqrt{7}}{-2}=\sqrt{7}\)
\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}=\sqrt{\left(\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}+\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}\right)^2}\)
\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}+2\sqrt{\frac{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}}+\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)
\(C=\sqrt{\frac{\left(4+\sqrt{7}\right)^2}{16-7}+\frac{\left(4-\sqrt{7}\right)^2}{16-7}+2}\)
\(C=\sqrt{\frac{\left(4+\sqrt{7}+4-\sqrt{7}\right)^2-2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{16-7}+2}\)
\(C=\sqrt{\frac{16^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{238}{9}+2}=\sqrt{\frac{256}{9}}=\frac{16}{3}\)
Chúc bạn học tốt ~
7 tháng 9 2016
a)\(\sqrt{1-2\sqrt{10}+10}=\sqrt{\left(1-\sqrt{10}\right)^2}=\left|1-\sqrt{10}\right|=\sqrt{10}-1\)
(vì 1<\(\sqrt{10}\))
b)\(\Rightarrow\sqrt{2}\left[\left(\sqrt{4-\sqrt{7}}\right)-\left(\sqrt{4+\sqrt{7}}\right)\right]=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}=\sqrt{7}-1-1-\sqrt{7}=-2\Rightarrow\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
VT
29 tháng 6 2018
A= \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\)\(1+\sqrt{7}+\sqrt{7}-1=2\sqrt{7}\)
\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
=\(\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\)\(\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)
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
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.
NH
4
NN
17 tháng 9 2020
a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
\(=\sqrt{7+4\sqrt{7}+4}-\sqrt{7-4\sqrt{7}+4}\)
\(=\sqrt{\left(\sqrt{7}+2\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}\)
\(=\left|\sqrt{7}+2\right|-\left|\sqrt{7}-2\right|\)
\(=\sqrt{7}+2-\sqrt{7}+2=4\)
17 tháng 9 2020
a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{\left(2+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}=2+\sqrt{7}-\sqrt{7}+2=4\)
b) \(A=\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)
\(\Rightarrow A^2=11-4\sqrt{6}-2\sqrt{\left(11-4\sqrt{6}\right)\left(11+4\sqrt{6}\right)}+11+4\sqrt{6}\)
\(A^2=22-2\sqrt{121-96}\)
\(A^2=22-2\sqrt{25}=22-2.5=12\)
\(\Rightarrow A=-\sqrt{12}\)(Chú ý \(A< 0\))
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 \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=8-6=2\)
\(\Rightarrow B=\sqrt{2}\)
24 tháng 7 2020
A = \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
A = \(\sqrt{2}+1-\sqrt{2}+1\)
A = 2
24 tháng 7 2020
B = \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
B = \(2-\sqrt{3}+\sqrt{3}+2\)
B = 4
1 tháng 10 2017
\(A=\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{15}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
\(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{4+\sqrt{7}}\sqrt{4-\sqrt{7}}-\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}-\left|4-\sqrt{7}\right|\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{16-7}-4+\sqrt{7}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=3-4+\sqrt{7}=-1+\sqrt{7}\)
\(\Leftrightarrow B=\frac{-1+\sqrt{7}}{\sqrt{4-\sqrt{7}}}\)
tíck mình nha bn thanks !!!!!!!!!!
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{7-2\sqrt{7}+1}+\sqrt{7+2\sqrt{7}+1}}{\sqrt{2}}\)
\(=\frac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
Đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(\sqrt{2}A=\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7}+1}-\sqrt{\left(\sqrt{7}\right)^2+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)