Bài `1`:
a) `3\sqrt{3}+4\sqrt{12}-5\sqrt{27}`
`=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}`
`=-4\sqrt{3}`
b) `1/2\sqrt{48}-2\sqrt{75}-(\sqrt{33})/(\sqrt{11})+5\sqrt{1(1)/(3)}`
`=2\sqrt{3}-10\sqrt{3}-\sqrt{(33)/(11)}+5\sqrt{4/3}`
`=-8\sqrt{3}-\sqrt{3}+(10\sqrt{3})/(3)`
`=-9\sqrt{3}+(10\sqrt{3})/(3)`
`=(-27\sqrt{3}+10\sqrt{3})/(3)`
`=(-17\sqrt{3})/(3)`
c)
`A=\sqrt{(1-\sqrt{3})^2}-\sqrt{(\sqrt{3}+2)^2}`
`A=|1-\sqrt{3}|-|\sqrt{3}+2|`
`A=1-\sqrt{3}-\sqrt{3}-2`
`A=-1`
d)
`B=\sqrt{(2-\sqrt{3})^2}+\sqrt{4-2\sqrt{3}}`
`B=|2-\sqrt{3}|+\sqrt{(1-\sqrt{3})^2}`
`B=2-\sqrt{3}+|1-\sqrt{3}|`
`B=2-\sqrt{3}+\sqrt{3}-1`
`B=1`
Bài `2`:
a) `\sqrt{1-4x+4x^2}=5` `(x∈R)`
⇔`\sqrt{(1-2x)^2}=5`
⇔`|1-2x|=5`
⇔\(\left[ \begin{array}{l}1-2x=5\\2x-1=5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2(tm)\\x=3(tm)\end{array} \right.\)
Vậy S={-2,3}`
b) `\sqrt{4-5x}=12` `(x<=4/5)`
⇔`4-5x=144`
⇔`-5x=140`
⇔`x=-28` `(tm)`
Vậy `S={-28}`
c) `\sqrt{4x+20}-3\sqrt{5+x}+4/3\sqrt{9x+45}=6` `(x>=-5)`
⇔`\sqrt{4(x+5)}-3\sqrt{x+5}+4/3\sqrt{9(x+5)}=6`
⇔`2\sqrt{5}-3\sqrt{5}+4\sqrt{x+5}=6`
⇔`3\sqrt{x+5}=6`
⇔`\sqrt{x+5}=2`
⇔`x+5=4`
⇔`x=-1` `(tm)`
Vậy `S={-1}`
