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25 tháng 7 2022

\(7-4\sqrt{3}=7-2.2\sqrt{3}=4-2.2\sqrt{3}+3=\left(2-\sqrt{3}\right)^2=\left(\sqrt{3}-2\right)^2\)

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

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

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

5 tháng 12 2019

\(a)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

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

\(=\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\)

\(=\frac{24}{2}\)

\(=12\)

\(b)\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2+\sqrt{8}-\sqrt{6}\right)}{2+\sqrt{2}-\sqrt{3}}\)

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

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

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

\(c)A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-3}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)

\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{44\left(2-\sqrt{3}\right)}{22}}\)

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

\(A=\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)

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

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

\(A=3-1=2\)

P/s: nếu đề là vậy thì t ra kết quả như vậy ạ, nhưng lần sau khi đăng câu hỏi bạn nên viết rõ hơn ra nhé

a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)

b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)

12 tháng 7 2019

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

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

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

\(=2\sqrt{3}\)

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

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

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

\(=2\sqrt{2}\)

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

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

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

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

\(=6\)

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

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

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

\(=4\)

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

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

\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)

\(=1+\sqrt{5}-\sqrt{5}+1\)

\(=2\)

12 tháng 7 2019

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

A = 2 + 2

A = 4

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

B = 3 + 3

B = 6

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

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

C = 3 + 3

C = 6

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

D = 2 + 2

D = 4

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

\(E=\sqrt{5}+1-\sqrt{5}+1\)

E = 1 + 1

E = 2

21 tháng 9 2017

14dm5cm=14,5dm;3dm7cm=3,7dm

chu vi hình chữ nhật đó là:

(14,5+3,7)x2=36,4(dm)

ĐS:36,4dm

21 tháng 9 2017

14 dm 5 cm = 14,5 dm 

3 dm 7 cm = 3,7 dm 

Chiều rộng HCN là :

14,5 - 3,7 = 10,8 ( dm )

chu vi HCN là :

( 14,5 + 10,8 ) x 2 = 50,6 ( dm )

ĐS:..

23 tháng 9 2017

a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

nhân cả hai vế với \(\sqrt{2}\), ta được:

\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)

\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)

\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)

\(=\sqrt{7}-1-\sqrt{7}-1\)

\(=-2\)

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

12 tháng 5 2018

a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)

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

23 tháng 5 2017

\(a,\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)

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

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

23 tháng 5 2017

\(b,\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{25}=5\)

23 tháng 6 2016

Hihi mình cũng học lớp 9, để mình giúp cậu nha!

a) \(\sqrt{3-2\sqrt{2}}+\sqrt{9+4\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(1+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{2}-1\right|+\left|1+\sqrt{8}\right|=\sqrt{2}-1+1+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)

b) \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}\)

\(=\left|\sqrt{7}-1\right|-\left|1+\sqrt{7}\right|=\sqrt{7}-1-1-\sqrt{7}=-2\)

c) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

(Nhớ click cho mình với nhoa!)