`a)\sqrt{9-4sqrt5}-sqrt5`

`=sqrt{5-2.2sqrt5+4}-sqrt5`

`=sqrt{(sqrt5-2)^2}-sqrt5`

`=|\sqrt5-2|-sqrt5`

`=sqrt5-2-sqrt5=-2`

`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`

`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`

`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`

`=|2-sqrt3|+|sqrt3-1|`

`=2-sqrt3+sqrt3-1=1`

`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`

`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`

`=sqrtx+7`

`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`

`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`

`=sqrt3+1-2sqrt3-1=-sqrt3`

`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)

a: \(=2\sqrt{2}+1-3=2\sqrt{2}-2\)

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

c: \(=2-\sqrt{3}+\sqrt{3}-1=1\)

\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)

b: \(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

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

c: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)

d: \(=\dfrac{\sqrt{18-2\sqrt{17}}-\sqrt{18+2\sqrt{17}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{17}-1-\sqrt{17}-1}{\sqrt{2}}=-\sqrt{2}\)

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương

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

\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)

\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)

\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)

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

=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)

CÂU CUỐI chưa làm đc

ý cuối cùng này :

\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có

\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)

\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)

\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)

=> \(A^2=13+\sqrt{7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7+....}}}}}\)

=>\(\left(A^2-13\right)^2=7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7...}}}}\)

=>\(\left(A^2-13\right)^2=7+A\)

Đến đây tách ra giải PT bậc 4 nha!

Bạn giải giúp mình với bấm không ra nghiệm nơi

a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)

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

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

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

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

\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)

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

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

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

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

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

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

d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)

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

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

e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)

f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)

e)