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VM
2
27 tháng 7 2018
I don't now
sorry
...................
nha
TB
28 tháng 7 2018
\(\frac{\sqrt{4+2\sqrt{3}}+2}{4+2\sqrt{3}-2\sqrt{4+2\sqrt{3}}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{4+2\sqrt{3}-2\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{\sqrt{3}+1+2}{4+2\sqrt{3}-2\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}+3}{4+2\sqrt{3}-2\sqrt{3}-2}\)
\(=\frac{\sqrt{3}+3}{2}\)
NT
1
H9
HT.Phong (9A5)
CTVHS
17 tháng 9 2023
\(A=\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}\right)^2-\left(\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)^2\)
\(A=\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}+\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left[\dfrac{2\left(2-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\dfrac{2\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right]\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left[\dfrac{2\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}\right)+2\left(2\sqrt{3}-2+3-\sqrt{3}\right)}{3-1}\right]\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}+2\sqrt{3}+1-\sqrt{3}\right)\left(\dfrac{4-2\sqrt{5}}{\sqrt{3}-1}-\dfrac{4+2\sqrt{3}}{\sqrt{3}+1}\right)\)
\(A=\left(3\sqrt{3}+3-\sqrt{15}-\sqrt{5}\right)\left[\dfrac{2\left(2\sqrt{3}+2-\sqrt{15}-\sqrt{5}-2\sqrt{3}+2-3+\sqrt{3}\right)}{2}\right]\)
\(A=\left(3\sqrt{3}+3-\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1-\sqrt{15}-\sqrt{5}\right)\)
\(A=\left(\sqrt{3}+1\right)^2\left(3-\sqrt{5}\right)\left(1-\sqrt{5}\right)\)
HG
0
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}}{3}\)
b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)
\(=2-\sqrt{5}-2\)
\(=-\sqrt{5}\)
NA
30 tháng 9 2018
\(A=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(A=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-\sqrt{3}+1}\)
\(A=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(A=\frac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)+\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(A=\frac{2\sqrt{3}-2+3-\sqrt{3}+2\sqrt{3}+2-3-\sqrt{3}}{\sqrt{3}\left(3-1\right)}\)
\(A=\frac{2\sqrt{3}}{2\sqrt{3}}=1\)
30 tháng 9 2018
\(A=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}.\)
\(=\frac{\left(2+\sqrt{3}\right)\left(2-\sqrt{4+2\sqrt{3}}\right)}{\left(2+\sqrt{4+2\sqrt{3}}\right)\left(2-\sqrt{4+2\sqrt{3}}\right)}+\frac{\left(2-\sqrt{3}\right)\left(2+\sqrt{4-2\sqrt{3}}\right)}{\left(2-\sqrt{4-2\sqrt{3}}\right)\left(2+\sqrt{4-2\sqrt{3}}\right)}\)
\(=\frac{4-2\sqrt{4+2\sqrt{3}}+2\sqrt{3}-\sqrt{3\left(4+2\sqrt{3}\right)}}{4-4-2\sqrt{3}}+\frac{4+2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}}{4-4+2\sqrt{3}}\)
\(=\frac{4+2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}}{2\sqrt{3}}-\frac{4-2\sqrt{4+2\sqrt{3}}+2\sqrt{3}-\sqrt{3\left(4+2\sqrt{3}\right)}}{2\sqrt{3}}\)
\(=\frac{4+2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}-4+2\sqrt{4+2\sqrt{3}}-2\sqrt{3}+\sqrt{3\left(4+2\sqrt{3}\right)}}{2\sqrt{3}}\)
\(=\frac{2\sqrt{4-2\sqrt{3}}-2\sqrt{3}-\sqrt{3\left(4-2\sqrt{3}\right)}+2\sqrt{4+2\sqrt{3}}-2\sqrt{3}+\sqrt{3\left(4+2\sqrt{3}\right)}}{2\sqrt{3}}\)
YY
5 tháng 7 2018
\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(\left(\sqrt{3}+1-\left(\sqrt{3}-1\right)\right)\left(\sqrt{3}+1+\sqrt{3}-1\right)=2\cdot2\sqrt{3}=4\sqrt{3}\)
cách 2 :\(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\right)\)
\(=\left(\sqrt{4+2\sqrt{3}}\right)^2-\left(\sqrt{4-2\sqrt{3}}\right)^2=4+2\sqrt{3}-\left(4-2\sqrt{3}\right)=4\sqrt{3}\)
15 tháng 9 2021
\(\left(\sqrt{3}+4\right)^2+\left(\sqrt{3}-1\right)^2\)
\(=3+8\sqrt{3}+16+3-2\sqrt{3}+1\)
\(=\left(3+16+3+1\right)+\left(8\sqrt{3}-2\sqrt{3}\right)\)
\(=23+6\sqrt{3}\)
22 tháng 10 2021
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)
3 2 + 4 2 = 9 + 16 = 25 = 5 2 = 5