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GV Nguyễn Trần Thành Đạt
Giáo viên
21 tháng 9 2023
\(a,\left(2\sqrt{3}-\sqrt{2}\right)^2+2\sqrt{24}=\left[\left(2\sqrt{3}\right)^2-2.2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\right]+2\sqrt{24}\\ =\left[12-4\sqrt{6}+2\right]+2\sqrt{24}=14-4\sqrt{6}+4\sqrt{6}=14\\ b,\left(3\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+2\sqrt{3}\right)-\sqrt{60}=3\sqrt{5}.\sqrt{5}-2\sqrt{3}.\sqrt{3}+3\sqrt{5}.2\sqrt{3}-\sqrt{3}.\sqrt{5}-\sqrt{60}\\ =15-6+6\sqrt{15}-\sqrt{15}-\sqrt{2^2.15}\\ =9+3\sqrt{15}\)
15 tháng 8 2018
\(A=\left(\sqrt{8}-3\sqrt{2}+10\right)\left(\sqrt{2}-3\sqrt{0.4}\right)=\sqrt{16}-\frac{12\sqrt{5}}{5}+\sqrt{20}-6\sqrt{10}-6+\frac{18\sqrt{5}}{5}\)
\(A=-2+\frac{16\sqrt{5}}{5}-6\sqrt{10}\)
b)\(B=\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{6+2\sqrt{5}}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2}-\frac{\sqrt{5}-1}{2}=1\)
18 tháng 8 2021
Bài 2:
a: Ta có: \(\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{60}+6\right):2\sqrt{3}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{12}\left(\sqrt{5}+\sqrt{3}\right):2\sqrt{3}\)
\(=2\sqrt{12}:2\sqrt{3}\)
=2
b: Ta có: \(\sqrt{5-\sqrt{21}}-\sqrt{\dfrac{7}{2}}\)
\(=\dfrac{\sqrt{10-2\sqrt{21}}-\sqrt{7}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-\sqrt{3}-\sqrt{7}}{\sqrt{2}}\)
\(=-\dfrac{\sqrt{6}}{2}\)
2 tháng 7 2021
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
2 tháng 7 2021
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
19 tháng 6 2017
2\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(14-\sqrt{84}+7-\sqrt{84}\)
= 21
16 tháng 6 2017
A\(\left(3-\sqrt{3}\right)\left(-2\sqrt{3}\right)+\left(3\sqrt{3}+1\right)^2\)=\(6-6\sqrt{3}+9+6\sqrt{3}+1\)
=16
B,\(\left(3\sqrt{5}-2\sqrt{3}\right)\sqrt{5}+\sqrt{60}\) =\(15-2\sqrt{15}+2\sqrt{15}=15\)
30 tháng 7 2021
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
AT
12 tháng 7 2021
1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)
2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)
\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}=\left|\sqrt{5}+\sqrt{3}\right|+3\sqrt{5}-2\sqrt{3}\)
\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)
4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
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