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12 tháng 8 2020
1)
BT = \(\sqrt{\left(5+\sqrt{3}\right)^2}-\sqrt{\left(5-\sqrt{3}\right)^2}\)
= \(\left(5+\sqrt{3}\right)-\left(5-\sqrt{3}\right)\)
= \(2\sqrt{3}\)
2:
BT = \(\frac{\sqrt{2}.\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\frac{5\left(\sqrt{6}-1\right)}{\left(1+\sqrt{6}\right)\left(\sqrt{6}-1\right)}\)
= \(\sqrt{6}-\frac{5\left(\sqrt{6}-1\right)}{5}\) = \(\sqrt{6}-\left(\sqrt{6}-1\right)\) =1
L
1
4 tháng 9 2023
\(\sqrt{28-10\sqrt{3}}\\ =\sqrt{3-10\sqrt{3}+25}\\ =\sqrt{\left(\sqrt{3}-5\right)^2}\\ =\left|\sqrt{3}-5\right|\\ =5-\sqrt{3}\)
\(\sqrt{41+12\sqrt{5}}\\ =\sqrt{5+12\sqrt{5}+36}\\ =\sqrt{\left(\sqrt{5}+6\right)}\\ =\left|\sqrt{5}+6\right|\\ =\sqrt{5}+6\)
\(\sqrt{32-10\sqrt{7}}\\ =\sqrt{7-10\sqrt{7}+25}\\ =\sqrt{\left(\sqrt{7}-5\right)^2}\\ =\left|\sqrt{7}-5\right|\\ =5-\sqrt{7}\)
\(\sqrt{11-4\sqrt{7}}\\ =\sqrt{7-4\sqrt{7}+4}\\ =\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|\\ =\sqrt{7}-2\)
FE
0
1 tháng 10 2021
a: Ta có: \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}-2\sqrt{5}\)
\(=-\sqrt{7}-2\sqrt{5}\)
DD
0
17 tháng 10 2020
\(\sqrt{28+10\sqrt{3}}\)=\(\sqrt{\left(\sqrt{3}\right)^2+2.5.\sqrt{3}+\left(\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{25}.\sqrt{3}+\left(\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+5\right)^2}\)
=\(|\sqrt{3}+5|\)
=\(\sqrt{3}+5\)(vì \(\sqrt{3}+5\)\(\ge0\))
GH
22 tháng 6 2023
a)
\(7\sqrt{2}=\sqrt{49.2}=\sqrt{98}\\ 2\sqrt{8}=\sqrt{4.8}=\sqrt{32}\\ 5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
Do 98 > 50 > 32 > 28 nên \(\sqrt{98}>\sqrt{50}>\sqrt{32}>\sqrt{28}\)
=> \(7\sqrt{2}>5\sqrt{2}>2\sqrt{8}>\sqrt{28}\)
b)
\(3\sqrt{10}=\sqrt{9.10}=\sqrt{90}\\ 5\sqrt{3}=\sqrt{25.3}=\sqrt{75}\)
\(\dfrac{20}{\sqrt{5}}=\dfrac{20\sqrt{5}}{5}=4\sqrt{5}=\sqrt{16.5}=\sqrt{80}\)
\(12\sqrt{\dfrac{2}{3}}=\sqrt{144.\dfrac{2}{3}}=\sqrt{96}\)
Do 96 > 90 > 80 > 75 => \(\sqrt{96}>\sqrt{90}>\sqrt{80}>\sqrt{75}\)
=> \(12\sqrt{\dfrac{2}{3}}>3\sqrt{10}>\dfrac{20}{\sqrt{5}}>5\sqrt{3}\)
17 tháng 7 2019
\(\left(\frac{15}{3-\sqrt{3}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
\(=\left(\frac{15}{3-\sqrt{3}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{3}+5\)
\(=\left(\frac{5\sqrt{3}}{\sqrt{3}-1}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{3}+5\)
\(=-\frac{5\sqrt{3}-8}{5-3\sqrt{3}}:\sqrt{3}+5\)
\(=-\frac{5\sqrt{3}-8}{\left(5-3\sqrt{3}\right)\left(5+3\sqrt{3}\right)}\)
\(=-\frac{5\sqrt{3}-8}{16-10\sqrt{3}}\)
= -(-1/2) = 1/2
FE
0
\(\sqrt{28-10\sqrt{3}}+\sqrt{28+10\sqrt{3}}=\sqrt{25-2.5.\sqrt{3}+3}+\sqrt{25+2.5.\sqrt{3}+3}\)
\(=\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{5^2+2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(5-\sqrt{3}\right)^2}+\sqrt{\left(5+\sqrt{3}\right)^2}\)
\(=\left|5-\sqrt{3}\right|+\left|5+\sqrt{3}\right|=5-\sqrt{3}+5+\sqrt{3}=10\)