Đáp án:
`a)` `B=1/{\sqrt{x}-5}` với `x\ge 0;x\ne 25`
`b)` `x\in {0;16;36;100}`
`c)` `x\in {1;9}`
Giải thích các bước giải:
`a)` `B=3/{\sqrt{x}+5}-{20-2\sqrt{x}}/{25-x} ` `(x\ge 0;x\ne 25)`
`=3/{\sqrt{x}+5}+{20-2\sqrt{x}}/{x-25}`
`={3(\sqrt{x}-5)+20-2\sqrt{x}}/{(\sqrt{x}+5)(\sqrt{x}-5)}`
`={\sqrt{x}+5}/{(\sqrt{x}+5)(\sqrt{x}-5)}`
`=1/{\sqrt{x}-5}`
Vậy `B=1/{\sqrt{x}-5}` với `x\ge 0;x\ne 25`
$\\$
`b)` `P=B\sqrt{x}=\sqrt{x}/{\sqrt{x}-5}``\quad (x\ge 0;x\ne 25)`
`={\sqrt{x}-5+5}/{\sqrt{x}-5}`
`={\sqrt{x}-5}/{\sqrt{x}-5}+5/{\sqrt{x}-5}`
`=1+5/{\sqrt{x}-5}`
Vì `1\in ZZ=>` Để `P\in ZZ` thì `5/{\sqrt{x}-5}\in ZZ`
`=>(\sqrt{x}-5)\in Ư(5)={-5;-1;1;5}`
`=>\sqrt{x}\in {0;4;6;10}`
`=>x\in {0;16;36;100}` (thỏa mãn)
Vậy `x\in {0;16;36;100}` thì `P` nguyên
$\\$
`c)` `A=|x-4|B`
`<=>{\sqrt{x}+2}/{\sqrt{x}-5}=|x-4|/{\sqrt{x}-5}` `(x\ge 0;x\ne 25)`
`<=>\sqrt{x}+2=|x-4|`
`<=>\sqrt{x}+2=|\sqrt{x}+2|.|\sqrt{x}-2|`
`<=>\sqrt{x}+2=(\sqrt{x}+2).|\sqrt{x}-2|`
(vì `\sqrt{x}+2\ge 2>0` với mọi `x\ge 0;x\ne 25`)
`<=>1=|\sqrt{x}-2|`
`<=>`$\left[\begin{array}{l}\sqrt{x}-2=1\\\sqrt{x}-2=-1\end{array}\right.$`<=>`$\left[\begin{array}{l}\sqrt{x}=3\\\sqrt{x}=1\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=9\\x=1\end{array}\right.\ (thỏa\ mãn)$
Vậy `x\in {1;9}` thỏa mãn đề bài
