a) Ta có: \(9+4\sqrt{5}\)

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

\(=\left(\sqrt{5}+2\right)^2\)(đpcm)

b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

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

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

=-2(ddpcm)

c) Ta có: \(\left(4-\sqrt{7}\right)^2\)

\(=16-2\cdot4\cdot\sqrt{7}+7\)

\(=23-8\sqrt{7}\)(đpcm)

d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)

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

\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)

\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)

\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

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

\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)

\(d.\) 

Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)

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

\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)

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

\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)

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

\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)

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

\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)

----------------------------

\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)

\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)

\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)

\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)

Cách 1:

\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)

\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)

Cách 2:

\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)

\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)

Vì $E>0$ nên $E=2$

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

Ta có

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

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

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

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

\(=-2=VP\left(đpcm\right)\)

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

Ta có:

\(VT=\frac{\sqrt{2}+1}{\sqrt{2}-1}\)

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

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

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

\(=3+2\sqrt{2}=VP\left(đpcm\right)\)

c,Bạn xem lại đề

\(d,\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}=8\)

Ta có:

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

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

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

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

\(=\frac{4+2\sqrt{5}-2\sqrt{5}+4}{\sqrt{5}^2-2^2}\)

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

\(=8=VP\left(đpcm\right)\)

\(e,\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)

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

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

\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}\)

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

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

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

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

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

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

\(=[3^2-\left(\sqrt{5}\right)^2].2\)

\(=4.2=8=VP\left(đpcm\right)\)

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

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

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

\(=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)( Vì \(2-\sqrt{3}>0\))

\(=4-2=1\)

mk

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

Câu a thì c/m được câu b đề yêu cầu gì thế.

a) Xét VP được :

\(\left(\sqrt{5}+2\right)^2\) sử dụng hàng đẳng thức số 1 :

\(\left(\sqrt{5}+2\right)^2=\sqrt{5}^2+2\cdot\sqrt{5}\cdot2+2^2=5+4\sqrt{5}+4=9+4\sqrt{5}=VT\)

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

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

Ta biến đổi vế phải :

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

=> Ta có VT= VP <=> VP = VT

b) Thiếu đề =.= sao làm

b,

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

\(=\sqrt{4-2.2\sqrt{5}+5}-\sqrt{5}\)

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

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

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

mấy bác tranh câu a e làm câu b