`B` `=`$\dfrac{x-4\sqrt{x}+5}{\sqrt{x}-2}$
`=` $\dfrac{x-4\sqrt{x}+4+1}{\sqrt{x}-2}$ `=` $\dfrac{(\sqrt{x}-2)^2+1}{$\sqrt{x}-2}$ = $\sqrt{x}$ `-` `2` `+` $\dfrac{1}{\sqrt{x}-2}$
`a)` `x` `=` $\sqrt{7-4\sqrt{3}}$ `+` $\sqrt{4-2\sqrt{3}}$ `=` $\sqrt{3-4\sqrt{3}+4}$+$\sqrt{3-2\sqrt{3}+1}$
`=` $\sqrt{(\sqrt{3}-2)^2}$ `+` $\sqrt{(\sqrt{3}-1)^2}$
`=` `2` `-` $\sqrt{3}$ `+` `\sqrt{3}` `-` `1`
`=` `1`
Thay `x=1` vào `B` ta có : $\sqrt{1}$ `-` `2` `+` $\dfrac{1}{\sqrt{1}-2}$ `=` `-2`
Vậy với `x =`$\sqrt{7-4\sqrt{3}}$+$\sqrt{4-2\sqrt{3}}$ thì `B` `=` `-2`
b) $x=\sqrt{61+ 28\sqrt{3}} -\sqrt{21+12\sqrt{3}}$
$=\sqrt{49+ 2.7.2\sqrt{3}+ 12}$ `-` $\sqrt{9+2.3.2\sqrt{3}+12}$
$=\sqrt{7^2+ 2.7.2\sqrt{3}+ (2\sqrt{3})^2}$ `-` $\sqrt{3^2+2.3.2\sqrt{3}+(2\sqrt{3})^2}$
$=\sqrt{\left(7+2\sqrt{3}\right)^2}$ `-` $\sqrt{\left(3+2\sqrt{3}\right)^2}$
`=`$7+2\sqrt{3}$ `-` $(3+2\sqrt{3})$
`=` `4`
Thay `x=4` vào `B` ta có : $\sqrt{4}$ `-` `2` `+` $\dfrac{1}{{\sqrt{4}-2}}$ `=` `0`
Vậy với $x=\sqrt{61+ 28\sqrt{3}} -\sqrt{21+12\sqrt{3}}$ thì `B=0`
Mình gửi phần a ,b ạ !!
Chúc bạn học tốt !!🙆
@Katniss
