CM : \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)

Giải :

VT= \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)

Thấy VT = VP = - 2

=> \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\) ( đpcm )

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)

\(VT=\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5-2}\right)^2}-\sqrt{5}=\left|\sqrt{5}-2\right|-\sqrt{5}\)

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

Ta thấy VT = VF = -2

\(\Rightarrow\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\left(đpcm\right)\)

Chúc bạn học tốt !!!

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)=\(\sqrt{\left(2\right)^2-2.2\sqrt{5}+5}-\sqrt{5}\)=\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)=\(|2-\sqrt{5}|-\sqrt{5}\)=\(\sqrt{5}-2-\sqrt{5}\)=\(-2\)=Vế Phải (điều phải chứng min)

\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)

\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)

\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)

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

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

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

\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

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

\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)

a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

Ta có : VT = \(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)

\(\Leftrightarrow VT=9\) \(=VP\)

Vậy.........

b) \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)

<=> \(\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2=6\)

Ta có : VT = \(2+\sqrt{3}+2-\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

= \(4+2\sqrt{4-3}=4+2=6\)

=> VT = VP

Vậy.....

c) \(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)

Ta có : VT = \(\dfrac{\sqrt{4}}{\sqrt{\left(2-\sqrt{5}\right)^2}}-\dfrac{\sqrt{4}}{\sqrt{\left(2+\sqrt{5}\right)^2}}\)

= \(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{2+\sqrt{5}}=\dfrac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

= \(\dfrac{8}{5-4}=8\)

=> VT = VP

Vậy....

a) Biến đổi vế trái ta có:

VT= \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)

= \(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)

= 9 = VP

Vậy đẳng thức đc chứng minh

b) Đặt vế trái = A = \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

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

\(A^2=4+2.\sqrt{4-3}=4+2.1=6\)

\(\Rightarrow A=\sqrt{6}=VP\)

Vậy đẳng thức đc chứng minh

a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

BĐVT ta có:

\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9=VP\)

Vậy đẳng thức đã được C/m

b) \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)

BĐVT ta có:

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

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

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

\(=\dfrac{\left|\sqrt{3+1}\right|+\left|\sqrt{3-1}\right|}{\sqrt{2}}=\dfrac{\sqrt{3+1}+\sqrt{3-1}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)

Vậy đẳng thức đã được C/m

c) \(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)

BĐVT ta có:

\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\sqrt{\dfrac{2^2}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{2^2}{\left(2+\sqrt{5}\right)^2}}\)

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

\(=\dfrac{2\sqrt{5}+4-2\sqrt{5}+4}{5-4}=8=VP\)

Vậy đẳng thức đã dược C/m