\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

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

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

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

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

\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

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

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

\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)

\(A=-1\)

\(a.\frac{2}{5}+\frac{3}{6}=\frac{2}{5}+\frac{1}{2}=\frac{4}{10}+\frac{5}{10}=\frac{9}{10}.\)

\(b.\frac{21}{30}-\frac{9}{30}=\frac{21-9}{30}=\frac{12}{30}=\frac{2}{5}\)

\(c.\frac{5}{8}\cdot\frac{4}{7}=\frac{5\cdot4}{8\cdot7}=\frac{5\cdot1}{2\cdot7}=\frac{5}{14}\)

\(d.\frac{7}{9}:\frac{11}{6}=\frac{7}{9}\cdot\frac{6}{11}=\frac{7\cdot6}{9\cdot11}=\frac{7\cdot2}{3\cdot11}=\frac{14}{33}\)

`A=\sqrt{6-2\sqrt{5}}`

`A=\sqrt{(\sqrt{5}-1)^2}`

`A=\sqrt{5}-1`

_________

`B=\sqrt{4-\sqrt{12}}=\sqrt{4-2\sqrt{3}}`

`B=\sqrt{(\sqrt{3}-1)^2}`

`B=\sqrt{3}-1`

_________

`C=\sqrt{19-8\sqrt{3}}`

`C=\sqrt{(4-\sqrt{3})^2}`

`C=4-\sqrt{3}`

_________

`D=\sqrt{5-2\sqrt{6}}`

`D=\sqrt{(\sqrt{3}-\sqrt{2})^2}`

`D=\sqrt{3}-\sqrt{2}`

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

\(B=\sqrt{4-\sqrt{12}}=\sqrt{4-\sqrt{4.3}}=\sqrt{4-2\sqrt{3}}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

\(C=\sqrt{19-8\sqrt{3}}=\sqrt{19-2.4.\sqrt{3}}\sqrt{\sqrt{3}^2-2.4.\sqrt{3}+4^2}=\sqrt{\left(\sqrt{3}-4\right)^2}=\sqrt{3}-4\)

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

a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

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

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

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

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

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

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

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

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

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

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

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

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

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

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

a: =1/4+1/4=2/4=1/2

b: =3/5+3/5=6/5

c: =5/9+7/9=12/9=4/3

a, 1/4+1/4=2/4=1/4

b, 3/5+3/5=6/5

c, 5/9+7/9=12/9=4/3

a)1/4+1/4=2/4=1/4

b) 3/5+3/5=6/5

c) 5/9+7/9=12/9=4/3

a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`

`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`

`=10(12x^2+25)-(120x^2+49)`

`=120x^2+250-120x^2-49`

`=201`

b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`

Lời giải:
a. 

$(2x+5)^3-(2x-5)^3-(120x^2+49)$

$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$

$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$

$=10(12x^2+25)-(120x^2+49)=250-49=201$

b.

$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$

$=7(1-10x)$

a) Ta có: \(\left(2x+5\right)^3-\left(2x-5\right)^3-\left(120x^2+49\right)\)

\(=8x^3+60x^2+150x+125-8x^3+60x^2-150x+125-120x^2-49\)

\(=201\)

b) Ta có: \(\left(4-5x\right)^2-\left(3+5x\right)^2\)

\(=\left(4-5x-3-5x\right)\left(4-5x+3+5x\right)\)

\(=7\left(-10x+1\right)\)

\(=-70x+7\)