`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)

\(a,\sqrt{45.8}=\sqrt{360}=\sqrt{6^2.10}=6\sqrt{10}\)

\(b,\sqrt{2,5.14,4}=\sqrt{36}=\sqrt{6^2}=6\)

\(c,\sqrt{10.40}=\sqrt{400}=\sqrt{20^2}=20\)

\(d,\sqrt{52}.\sqrt{13}=\sqrt{52.13}=\sqrt{676}=\sqrt{26^2}=26\)

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

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

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

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

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

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

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

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

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

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

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

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

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) 9
b)90
c)8
d)7/10
e)2/5

a=9

b=90

c=2

d=0,7

e=0,2

Lời giải:

a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)

b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)

c.

\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)

\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)

\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)

d.

\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)

\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)

\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)

a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)

b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)

d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=5+3\sqrt{2}\)

a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)

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

c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)

`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`

`=2/sqrt2=sqrt2`

`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`

`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`

`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`

`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`

`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`

`=(-2sqrt3)/sqrt2=-sqrt6`

`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`

`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`

`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`

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

`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`

`=(2sqrt3)/sqrt2=sqrt6`

`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`

`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`

`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`

`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`

`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`

`=2/sqrt2=sqrt2`

a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

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

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

b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)

\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)

\(=-3\cdot4\cdot3\sqrt{10}\)

\(=-36\sqrt{10}\)

b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)

\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)

\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)

\(=15\sqrt{3}\)

c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)

\(=4\cdot3-4-2\cdot2\)

\(=12-4-4=4\)