Đáp án:
`A=2`
`B=5/(\sqrtx+3)`
Giải thích các bước giải:
$*$
`A=\sqrt{4+2\sqrt3}-2/(\sqrt3+1)`
`\to A=\sqrt{3+2\sqrt3+1}-2/(\sqrt3+1)`
`\to A=\sqrt{(\sqrt3+1)^2}-2/(\sqrt3+1)`
`\to A=|\sqrt3+1|-2/(\sqrt3+1)`
`\to A=(\sqrt3+1)-2/(\sqrt3+1)`
`\to A=((\sqrt3+1)^2-2)/(\sqrt3+1)`
`\to A=(3+2\sqrt3+1-2)/(\sqrt3+1)`
`\to A=(2+2\sqrt3)/(\sqrt3+1)`
`\to A=(2(\sqrt3+1))/(\sqrt3+1)`
`\to A=2`
Vậy `A=2`
$*$
Với `x≥0;x\ne9`
Ta có:
`B=3/(\sqrtx+3)-2/(\sqrtx-3)+(4\sqrtx)/(x-9)`
`\to B=3/(\sqrtx+3)-2/(\sqrtx-3)+(4\sqrtx)/((\sqrtx-3)(\sqrtx+3))`
`\to B=(3(\sqrtx-3)-2(\sqrtx+3)+4\sqrtx)/((\sqrtx-3)(\sqrtx+3))`
`\to B=(3\sqrtx-9-2\sqrtx-6+4\sqrtx)/((\sqrtx-3)(\sqrtx+3))`
`\to B=(5\sqrtx-15)/((\sqrtx-3)(\sqrtx+3))`
`\to B=(5(\sqrtx-3))/((\sqrtx-3)(\sqrtx+3))`
`\to B=5/(\sqrtx+3)`
Vậy với `x≥0;x\ne9` thì `B=5/(\sqrtx+3)`
