Bài làm :
a, `\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}`
`=\sqrt{7+2.\sqrt{7}.1+1}-\sqrt{7-2.\sqrt{7}.1+1}`
`=\sqrt{(\sqrt{7})^2+2.\sqrt{7}.1+1^2}-\sqrt{(\sqrt{7})^2-2.\sqrt{7}.1+1^2}`
`=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}`
`=|\sqrt{7}+1|-|\sqrt{7}-1|`
`=\sqrt{7}+1-(\sqrt{7}-1)`
`=\sqrt{7}+1-\sqrt{7}+1`
`=2`
b, `\sqrt{13+4\sqrt{3}}+\sqrt{4-2\sqrt{3}}-3\sqrt{3}`
`=\sqrt{12+2.2\sqrt{3}.1+1}+\sqrt{3-2\sqrt{3}.1+1}-3\sqrt{3}`
`=\sqrt{(2\sqrt{3})^2+2.2\sqrt{3}.1+1^2}+\sqrt{(\sqrt{3})^2-2\sqrt{3}.1+1^2}-3\sqrt{3}`
`=\sqrt{(2\sqrt{3}+1)^2}+\sqrt{(\sqrt{3}-1)^2}-3\sqrt{3}`
`=|2\sqrt{3}+1|+|\sqrt{3}-1|-3\sqrt{3}`
`=2\sqrt{3}+1+\sqrt{3}-1-3\sqrt{3}`
`=0`
c, Đặt `A=\sqrt{8-\sqrt{15}}-\sqrt{8+\sqrt{15}}`
`=>\sqrt{2}.A=\sqrt{2}.\sqrt{8-\sqrt{15}}-\sqrt{2}.\sqrt{8+\sqrt{15}}`
`=>\sqrt{2}.A=\sqrt{16-2\sqrt{15}}-\sqrt{16+2\sqrt{15}}`
`=>\sqrt{2}.A=\sqrt{15-2\sqrt{15}.1+1}-\sqrt{15+2\sqrt{15}.1+1}`
`=>\sqrt{2}.A=\sqrt{(\sqrt{15})^2-2\sqrt{15}.1+1^2}-\sqrt{(\sqrt{15})^2+2\sqrt{15}.1+1^2}`
`=>\sqrt{2}.A=\sqrt{(\sqrt{15}-1)^2}-\sqrt{(\sqrt{15}+1)^2}`
`=>\sqrt{2}.A=|\sqrt{15}-1|-|\sqrt{15}+1|`
`=>\sqrt{2}.A=\sqrt{15}-1-(\sqrt{15}+1)`
`=>\sqrt{2}.A=\sqrt{15}-1-\sqrt{15}-1`
`=>\sqrt{2}.A=-2`
`=>A=-\sqrt{2}`
d, `\sqrt{a^2}-3a`
`=|a|-3a`
`=a-3a` ( vì `a>0` )
`=-2a`
e, `2x-2+\sqrt{4x^2-4x+1}`
`=2x-2+\sqrt{(2x-1)^2}`
`=2x-2+|2x-1|`
`=2x-2+2x-1` ( vì `x>1` )
`=4x-3`
f, `\sqrt{4x^2-4x+1}+2x`
`=\sqrt{(2x-1)^2}+2x`
`=|2x-1|+2x`
`=-(2x-1)+2x` ( vì `x<0,5` )
`=-2x+1+2x`
`=1`
