Đáp án:
a) `A= -\sqrt{7}`
`B=``\frac{8}{\sqrt{x}-3}`
b) `0<x<9`
Giải thích các bước giải:
a) `A=\sqrt{28}-\sqrt{63}+``\frac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{(\sqrt{7}+1)^2}`
`=2\sqrt{7}-3\sqrt{7}+``\frac{7+(\sqrt{7}+1)}{\sqrt{7}}-|\sqrt{7}+1|`
`=``-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}`
`B=``(\frac{1}{\sqrt{x}+3}+``\frac{1}{\sqrt{x}-3})``\frac{4\sqrt{x}+12}{\sqrt{x}}`
`=``\frac{\sqrt{x}-3+\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}.``\frac{4(\sqrt{x}+3)}{\sqrt{x}}`
`=``\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.``\frac{4(\sqrt{x}+3)}{\sqrt{x}}`
`=``\frac{8}{\sqrt{x}-3}`
b) `A>B =>``-\sqrt{7}>``\frac{8}{\sqrt{x}-3}=>``\frac{8}{\sqrt{x}-3}+\sqrt{7}<0`
`⇒` `\frac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}<0`
Ta có: $\begin{cases}8=\sqrt{64}\\ 3\sqrt{7}=\sqrt{63}\end{cases}$ `=>8-3\sqrt{7}>0=>8-3\sqrt{7}+\sqrt{7x}>0`
`=>\sqrt{x}-3<0=>\sqrt{x}<3=>x<9=>0<x<9`
