`A=(sqrt{x}+2)/(sqrt{x}-5) \qquad B=3/(sqrt{x}+5)+(20-2sqrt{x})/(x-25)`
`a)`ĐKXĐ: `x>=0;x\ne25`
`\qquad A>0`
`<=> (sqrt{x}+2)/(sqrt{x}-5)>0`
Do `sqrt{x}+2>0` với `AAx>=0`
`=> sqrt{x}-5>0`
`<=> sqrt{x}>5`
`<=> x>25`
Vậy `x>25` thì `A>0`
`b)` ĐKXĐ: `x>=0;x\ne25`
`B=3/(sqrt{x}+5)+(20-2sqrt{x})/(x-25)`
`B=(3(sqrt{x}-5)+20-2sqrt{x})/((sqrt{x}+5)(sqrt{x}-5))`
`B=(3sqrt{x}-15+20-2sqrt{x})/((sqrt{x}+5)(sqrt{x}-5))`
`B=(sqrt{x}+5)/((sqrt{x}+5)(sqrt{x}-5))`
`B=1/(sqrt{x}-5)`
Vậy `B=1/(sqrt{x}-5)` khi `x>=0;x\ne25`
`c) A=B.|x-4|`
`<=> (sqrt{x}+2)/(sqrt{x}-5)=1/(sqrt{x}-5). |x-4|`
`<=> (sqrt{x}+2)/(sqrt{x}-5)=|x-4|/(sqrt{x}-5)`
`=> sqrt{x}+2=|x-4|`
`<=> |(sqrt{x}+2)(sqrt{x}-2)|-(sqrt{x}+2)=0`
`<=> (sqrt{x}+2)(|sqrt{x}-2|-1)=0`
`=> |sqrt{x}-2|=1` $(\text{do $\sqrt{x}+2>0$ với mọi x})$
`<=> [(sqrt{x}-2=1),(sqrt{x}-2=-1):}<=>[(sqrt{x}=3),(sqrt{x}=1):}<=>[(x=9 (\text{tm})),(x=1(\text{tm})):}`
Vậy `x\in{1;9}` thì `A=B.|x-4|`
