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20 tháng 9 2021
a) \(=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}=2+\sqrt{3}-1-\sqrt{3}=1\)
c) \(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}\)
d) \(=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1\)
12 tháng 7 2021
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{2}}{2}\)
___________
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
__________
\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
14 tháng 8 2023
a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
7.
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\sqrt{4.3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(\sqrt{4}+\sqrt{3})^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-2.5\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
AH
Akai Haruma
Giáo viên
20 tháng 7 2020
5.
\(\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}\)
6.
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\)
\(=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{(\sqrt{2}+\sqrt{5})^2}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}-\sqrt{(\sqrt{2}+\sqrt{5})^2}=|\sqrt{2}+\sqrt{5}+1|-|\sqrt{2}+\sqrt{5}|=1\)
AH
13 tháng 8 2018
mình biết nội quy rồi nên đưng đăng nội quy
ai chơi bang bang 2 kết bạn với mình
mình có nick có 54k vàng đang góp mua pika
ai kết bạn mình cho
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 1 2022
Đặt \(\left\{{}\begin{matrix}\sqrt{\sqrt{2}+1}=a\\\sqrt{\sqrt{2}-1}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2\sqrt{2}=\sqrt{8}\)
\(\Rightarrow\sqrt{a^2+b^2}=\sqrt[4]{8}\)
Do đó:
\(A=\dfrac{\sqrt{\sqrt{a^2+b^2}+b}-\sqrt{\sqrt{a^2+b^2}-b}}{\sqrt{\sqrt{a^2+b^2}-a}}>0\)
\(\Rightarrow A^2=\dfrac{2\sqrt{a^2+b^2}-2\sqrt{a^2+b^2-b^2}}{\sqrt{a^2+b^2}-a}=\dfrac{2\left(\sqrt{a^2+b^2}-a\right)}{\sqrt{a^2+b^2}-a}=2\)
\(\Rightarrow A=\sqrt{2}\)
17 tháng 12 2023
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
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