Rút gọn biểu thức:
\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
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Những câu hỏi liên quan
NL
Nguyễn Lê Phước Thịnh
CTVHS
23 tháng 11 2025
Ta có: \(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\)
\(=\frac{-\left(\sqrt{a}+2\right)}{\sqrt{a}-2}+\frac{\sqrt{a}-2}{\sqrt{a}+2}-\frac{4a}{a-4}\)
\(=\frac{-\left(\sqrt{a}+2\right)^2+\left(\sqrt{a}-2\right)^2-4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{-a-4\sqrt{a}-4+a-4\sqrt{a}+4-4a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{-4a-8\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=-\frac{4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=-\frac{4\sqrt{a}}{\sqrt{a}-2}\)
Ta có: \(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\)
\(=\frac{-2}{\sqrt{a}-2}+\frac{\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-2\sqrt{a}+\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
Ta có: \(\left(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\right):\left(\frac{2}{2-\sqrt{a}}-\frac{\sqrt{a}+3}{2\sqrt{a}-a}\right)\)
\(=\frac{-4\sqrt{a}}{\sqrt{a}-2}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-4\sqrt{a}}{\sqrt{a}-2}\cdot\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+3}=\frac{4a}{\sqrt{a}-3}\)
VL
2
H
5 tháng 4 2021
\(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}=\sqrt{a}-\left(\sqrt{a}+2\right)=-2\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
5 tháng 4 2021
Ta có: \(P=\dfrac{a+2\sqrt{a}}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}-\sqrt{a}-2=-2\)
AH
Akai Haruma
Giáo viên
6 tháng 7 2019
Lời giải:
\(Q=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
21 tháng 6 2021
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{6-2+2\sqrt{3}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(1+\sqrt{3}\right)^2}=1+\sqrt{3}\)