Bài 1: Căn bậc hai

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nguyen thi khanh nguyen

Thực hiện các phép tính sau:

a,\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

b,\(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

c,\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

d,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

e,\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

f,\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

Giải giúp mình với m.nkhocroi mình đang cần gấp lắm!!!!!

Mysterious Person
14 tháng 7 2017 lúc 16:55

a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|=\left(\sqrt{5}-\sqrt{2}\right)-\left(\sqrt{5}+\sqrt{2}\right)\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}=-2\sqrt{2}\)

c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|=\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)\)

\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)

Mysterious Person
14 tháng 7 2017 lúc 17:15

d) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=2\sqrt{6+2\sqrt{5}}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=2\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{5}-2=2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

Mới vô
14 tháng 7 2017 lúc 17:32

e,

\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{4}\cdot\sqrt{2}}+\sqrt{9+2\cdot\sqrt{4}\cdot\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{8}}+\sqrt{9+2\cdot\sqrt{8}}\\ =\sqrt{9-2\cdot3\cdot\sqrt{8}+8}+\sqrt{8+2\cdot\sqrt{8}\cdot1+1}\\ =\sqrt{3^2-2\cdot3\cdot\sqrt{8}+\sqrt{8}^2}+\sqrt{\sqrt{8}^2+2\cdot\sqrt{8}\cdot1+1^2}\\ =\sqrt{\left(3-\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{8}+1\right)^2}\\ =\left|3-\sqrt{8}\right|+\left|\sqrt{8}+1\right|\\ =3-\sqrt{8}+\sqrt{8}+1\\ =4\)

f,

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{9}\cdot\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{18}}\\ =\sqrt{4-2\cdot2\cdot\sqrt{2}+2}+\sqrt{18-2\cdot2\cdot\sqrt{18}+4}\\ =\sqrt{2^2-2\cdot2\cdot\sqrt{2}+\sqrt{2}^2}+\sqrt{\sqrt{18}^2-2\cdot2\cdot\sqrt{18}+2^2}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{18}-2\right)^2}\\ =\left|2-\sqrt{2}\right|+\left|\sqrt{18}-2\right|\\ =2-\sqrt{2}+\sqrt{18}-2\\ =\sqrt{2}+\sqrt{18}\\ =\sqrt{2}+\sqrt{9}\cdot\sqrt{2}\\ =\sqrt{2}+3\sqrt{2}\\ =4\sqrt{2}\)

๖ۣۜTina Ss
7 tháng 12 2017 lúc 12:24

a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

Đặt \(A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(\Rightarrow\sqrt{2}A=\sqrt{2}\left(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\right)\)

\(\sqrt{2}A=\sqrt{10+4\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

\(\sqrt{2}A=\sqrt{\left(2+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)

\(\sqrt{2}A=\left|2+\sqrt{6}\right|-\left|\sqrt{6}-2\right|\)

\(\sqrt{2}A=\sqrt{6}+2-\sqrt{6}+2\)

\(\sqrt{2}A=4\)

\(\Rightarrow A=2\sqrt{2}\)

๖ۣۜTina Ss
7 tháng 12 2017 lúc 12:31

b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

Đặt \(B=\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(\Rightarrow\sqrt{2}B=\sqrt{2}\left(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\right)\)

\(=\sqrt{14-4\sqrt{10}}-\sqrt{14+4\sqrt{10}}\)

\(=\sqrt{\left(\sqrt{10}-2\right)^2}-\sqrt{\left(\sqrt{10}+2\right)^2}\)

\(=\left|\sqrt{10}-2\right|-\left|\sqrt{10}+2\right|\)

\(=\sqrt{10}-2-\sqrt{10}-2\)

\(=-4\)

\(\Rightarrow B=\dfrac{-4}{\sqrt{2}}=-2\sqrt{2}\)

๖ۣۜTina Ss
7 tháng 12 2017 lúc 12:35

c)\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}-1+\sqrt{3}+1\)

\(=2\sqrt{3}\)

๖ۣۜTina Ss
7 tháng 12 2017 lúc 12:40

f) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

\(=\left|2-\sqrt{2}\right|+\left|3\sqrt{2}-2\right|\)

\(=2-\sqrt{2}+3\sqrt{2}-2\)

\(=2\sqrt{2}\)

An Võ (leo)
18 tháng 6 2018 lúc 21:15

f ) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

=\(\sqrt{4-4\sqrt{2}+2}+\sqrt{22-4.\sqrt{9}\sqrt{2}}\)

= \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{18-4\sqrt{18}+4}\)

= \(\left|2-\sqrt{2}\right|+\sqrt{\left(\sqrt{18}-2\right)^2}\)

= \(2-\sqrt{2}+\left|3\sqrt{2}-2\right|\)

= \(2-\sqrt{2}+3\sqrt{2}-2\)

= \(2\sqrt{2}\)

tran nguyen bao quan
17 tháng 10 2018 lúc 20:30

Mình sẽ giải câu e và f

e) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.1+1}=\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.3+3^2}+\sqrt{\left(2\sqrt{2}\right)^2+2.2\sqrt{2}.1+1}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=\left|2\sqrt{2}-3\right|+\left|2\sqrt{2}+1\right|=3-2\sqrt{2}+2\sqrt{2}+1=4\)

f) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=\sqrt{4-2.2\sqrt{2}+2}+\sqrt{18-2.3\sqrt{2}.2+4}=\sqrt{2^2-2.2\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}\right)^2-2.3\sqrt{2}.2+2^2}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=\left|2-\sqrt{2}\right|+\left|3\sqrt{2}-2\right|=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)


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