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2:
ĐKXĐ: x>=3
\(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)
=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\sqrt{3}+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)
TH1: x>=6
(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)
=>\(2\sqrt{x-3}=2\sqrt{3}\)
=>x-3=3
=>x=6(nhận)
TH2: 3<=x<6
Phương trình (1) sẽ là;
\(\sqrt{x-3}+\sqrt{3}-\sqrt{x-3}=\sqrt{3}\)
=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)
1:
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)
\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)
\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)
\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)
\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)
=>\(A=\sqrt{10}+\sqrt{2}\)
Bài 1: Ta có:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{6-2}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{6+2-2\sqrt{6.2}}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{(\sqrt{6}-\sqrt{2})^2}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2}=\frac{6-2}{2}=2\)
Bài 2:
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)
\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)
\(=16+2\sqrt{64-4(10+2\sqrt{5})}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)
\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}\)
\(=16+2(\sqrt{20}-2)=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)
\(\Rightarrow A=\sqrt{10}+\sqrt{2}\)
đề là rút gọn các biểu thức sau
nhờ mọi người giải giúp mình. cảm ơn mn nhìu
a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)
=2căn 5-2-2căn 5
=-2
d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\) = \(\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{-\dfrac{4}{1+\sqrt{5}}}\) = \(2\sqrt{5}-2\left(1+\sqrt{5}\right)\) = -2.
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)
\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)
\(A^2=144-80-16\sqrt{5}\)
\(A^2=64-16\sqrt{5}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)
\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=2+2.\sqrt{2}.\sqrt{10}+10\)
\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)
=> \(A=\sqrt{2}+\sqrt{10}\)
A = \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}-\sqrt{2}-\sqrt{10}\)
Ta có : B = \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}-\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow B^2=16-2\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(=16-2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)
\(=16-2\sqrt{24-8\sqrt{5}}\)
\(=16-2\sqrt{\left(2\sqrt{5}-2\right)^2}=16-2\left(2\sqrt{5}-2\right)\)
\(=20-4\sqrt{5}\)
Vì \(8+2\sqrt{10+\sqrt{5}}>8-2\sqrt{10+2\sqrt{5}}\)
\(\Rightarrow B>0\)
\(\Rightarrow B=\sqrt{20-4\sqrt{5}}=2\sqrt{5-\sqrt{5}}\)
\(\Rightarrow A=B-\sqrt{2}-\sqrt{10}=2\sqrt{5-\sqrt{5}}-\sqrt{2}-\sqrt{10}=2\)
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath
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\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
=\(16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}\) =\(16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(\Rightarrow A=\sqrt{12+4\sqrt{5}}\)