Question

Answer 1

a) \(A=\sqrt{12+2\sqrt{11}}-\sqrt{11}=\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{11}=\sqrt{11}+1-\sqrt{11}=1\)

b) \(B=\sqrt{14-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}=3-\sqrt{5}+2\sqrt{5}-3=\sqrt{5}\)

c) \(C=\sqrt{y-4+2\sqrt{y-5}}-\sqrt{y-5}=\sqrt{\left(\sqrt{y-5}+1\right)^2}-\sqrt{y-5}=\sqrt{y-5}+1-\sqrt{y-5}=1\)

Answer 2

Lời giải:

a. $A=\sqrt{11+2\sqrt{11}+1}-\sqrt{11}$

$=\sqrt{(\sqrt{11}+1)^2}-\sqrt{11}$

$=|\sqrt{11}+1|-\sqrt{11}=\sqrt{11}+1-\sqrt{11}=1$

b.

$B=\sqrt{3^2-2.3\sqrt{5}+5}+\sqrt{20-2\sqrt{20}.\sqrt{9}+9}$

$=\sqrt{(3-\sqrt{5})^2}+\sqrt{(20-\sqrt{9})^2}$
$=|3-\sqrt{5}|+|2\sqrt{5}-3|$

$=3-\sqrt{5}+2\sqrt{5}-3=\sqrt{5}$

c.

$C=\sqrt{y-5)+2\sqrt{y-5}+1}-\sqrt{y-5}$

$=\sqrt{(\sqrt{y-5}+1)^2}-\sqrt{y-5}$

$=|\sqrt{y-5}+1|-\sqrt{y-5}=\sqrt{y-5}+1-\sqrt{y-5}=1$

 

Answer 3

a) \(A=\sqrt{12+2\sqrt{11}}-\sqrt{11}=\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{11}=\sqrt{11}+1-\sqrt{11}=1\)

 

Answer 4

c: Ta có: \(C=\sqrt{y-4+2\sqrt{y-5}}-\sqrt{y-5}\)

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

=1