\(\left(\sqrt{18}+\sqrt{20}-\sqrt{8}\right).\sqrt{2}-2\sqrt{10}\)
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S
28 tháng 7 2019
\(\left(\sqrt{125}-\sqrt{18}-\sqrt{5}-\sqrt{2}\right)\left(\sqrt{125}+2\sqrt{8}-\sqrt{20}-\sqrt{2}\right)=\text{[}\left(5\sqrt{5}-\sqrt{5}\right)-\left(3\sqrt{2}+\sqrt{2}\right)\text{]}\text{[}\left(5\sqrt{5}-2\sqrt{5}\right)+\left(4\sqrt{2}-\sqrt{2}\right)\text{]}=\left(4\sqrt{5}-4\sqrt{2}\right)\left(3\sqrt{5}-3\sqrt{2}\right)=12\left(\sqrt{5}-\sqrt{2}\right)^2=12\left(7-2\sqrt{10}\right)=84-24\sqrt{10}\)
28 tháng 7 2019
\(\left(\sqrt{125}-\sqrt{18}-\sqrt{5}-\sqrt{2}\right)\left(\sqrt{125}+2\sqrt{8}-\sqrt{20}-\sqrt{2}\right)\)
\(=\left(5\sqrt{5}-3\sqrt{2}-\sqrt{5}-\sqrt{2}\right)\left(5\sqrt{5}+4\sqrt{2}-2\sqrt{5}-\sqrt{2}\right)\)
\(=\left(4\sqrt{5}-4\sqrt{2}\right)\left(3\sqrt{5}+3\sqrt{2}\right)\)
\(=4\cdot\left(\sqrt{5}-\sqrt{2}\right)\cdot3\cdot\left(\sqrt{5}+\sqrt{2}\right)\)
\(=12\left(5-2\right)\)
\(=12\cdot3\)
\(=36\)
10 tháng 7 2021
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
10 tháng 7 2021
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
10 tháng 7 2021
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
10 tháng 7 2021
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 7 2021
\(A=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{2}.\sqrt{6-2\sqrt{5}}+\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{2}\left(\sqrt{5}-1\right)+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{10}+\sqrt{2}+\sqrt{10}-\sqrt{5}}{2\left(\sqrt{2}+1\right)}\)
\(=\dfrac{\sqrt{2}+1}{2\left(\sqrt{2}+1\right)}=\dfrac{1}{2}\)
NV
0
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: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
20 tháng 7 2017
c) \(\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\left(6-2\sqrt{5}\right)\)
= \(\left(\sqrt{5}+1\right)\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)\left(5-2\sqrt{5}+1\right)\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{6+2\sqrt{5}}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{5+2\sqrt{5}+1}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)\left(\sqrt{\left(\sqrt{5}+1\right)^2}\right)\left(\sqrt{5}-1\right)^2\)
= \(\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)
= \(4.4=16\)
d) \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{1+2+5+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{\left(\sqrt{2}+\sqrt{1}+\sqrt{5}\right)^2}-\sqrt{2}-\sqrt{5}\)
= \(\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}\)
= \(1\)
Bài làm:
Ta có: \(\left(\sqrt{18}+\sqrt{20}-\sqrt{8}\right).\sqrt{2}-2\sqrt{10}\)
\(=\left(3\sqrt{2}+2\sqrt{5}-2\sqrt{2}\right).\sqrt{2}-2\sqrt{10}\)
\(=\left(\sqrt{2}+2\sqrt{5}\right)-2\sqrt{10}\)
\(=2+2\sqrt{10}-2\sqrt{10}\)
\(=2\)