Chứng minh các hằng đẳng thức:
a) \(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
b) \(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}=\sqrt{2}+1\)
K
Khách
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Những câu hỏi liên quan
N
2
6 tháng 7 2021
\(10+\sqrt{60}+\sqrt{24}+\sqrt{40}=10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}\)
\(=\left(5+2\sqrt{15}+3\right)+2+2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2+2\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)+2\)
\(=\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{10+\sqrt{60}+\sqrt{24}+\sqrt{40}}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
6 tháng 7 2021
Dùng hẳng đẳng thức 3 số:
$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$
$VT=\sqrt{5+3+2+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}=\sqrt{(\sqrt5+\sqrt3+\sqrt2)^2}=VP(đpcm)$
TH
0
TH
1
18 tháng 9 2016
\(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}=\sqrt{3+2+1+\sqrt{2^2.2.3}+\sqrt{2^2.3}+\sqrt{2^2.2}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2+1^2+2\sqrt{3}.\sqrt{2}+2\sqrt{3}.1+2\sqrt{2}.1}=\sqrt{\left(\sqrt{3}+\sqrt{2}+1\right)^2}\)
(áp dụng hằng đẳng thức (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc)
\(=\sqrt{3}+\sqrt{2}+1\)
9 tháng 10 2021
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
TT
1
22 tháng 6 2017
\(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}\)
\(=\sqrt{2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\left(\sqrt{3}+\sqrt{5}-\sqrt{2}\right)^2\)
\(=\sqrt{3}+\sqrt{5}-\sqrt{2}\)
NT
0
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,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 6 2019
\(\sqrt{10+2\sqrt{15}+2\sqrt{6}+2\sqrt{10}}\)
\(=\sqrt{5+3+2+2\sqrt{5}.\sqrt{3}+2\sqrt{3}.\sqrt{2}+2\sqrt{5}.\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
Bạn áp dụng hằng đẳng thức (a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc)