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alibaba nguyễn
17 tháng 8 2016
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27 tháng 7 2023
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
26 tháng 10 2023
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)
\(=3+\sqrt{5}+3-\sqrt{5}=6\)
c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)
\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)
\(=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}\)
d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)
\(=\sqrt{16-3+3}=\sqrt{16}=4\)
e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
TP
2
TT
23 tháng 6 2023
\(A=\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)
\(=\left|2\sqrt{3}-3\sqrt{2}\right|+\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}+\sqrt{1^2}}-\sqrt{\left(3\sqrt{2}\right)^2+2.2.3\sqrt{2}+2^2}\)
\(=-2\sqrt{3}+3\sqrt{2}+\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(3\sqrt{2}+2\right)^2}\)
\(=-2\sqrt{3}+3\sqrt{2}+\left|2\sqrt{3}-1\right|-\left|3\sqrt{2}+2\right|\)
\(=-2\sqrt{3}+3\sqrt{2}+2\sqrt{3}-1-3\sqrt{2}-2\)
\(=-3\)
23 tháng 6 2023
\(A=3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-1-3\sqrt{2}-2=-3\)
18 tháng 7 2023
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)
\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)
\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)
\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)
\(=2\sqrt{6}-1\)
18 tháng 7 2023
\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)
9 tháng 5 2022
\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\left|\sqrt{5}+2\right|+\left|\sqrt{5}-2\right|\)
\(B=\sqrt{5}+2+\sqrt{5}-2\)
\(B=2\sqrt{5}\)
9 tháng 5 2022
\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)
\(A=-1\)
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
