Đáp án:
`1)` `A=3/2`
`2)` `B=1/(sqrtx+3)`
`3)` `x in {0;9}`
Giải thích các bước giải :
`1)` `x = 49=>A=(2sqrtx+1)/(sqrtx+3)=(2sqrt49+1)/(sqrt49+3)=(2*7+1)/(7+3)=15/10=3/2`
`2)` `B = (2/(sqrtx+3)-(sqrtx-5)/(x-9)):(sqrtx-1)/(sqrtx-3)(x>=0,xne1,xne9)` $\\$ `= (2/(sqrtx+3)-(sqrtx-5)/[(sqrtx+3)(sqrtx-3)])*(sqrtx-3)/(sqrtx-1)` $\\$ `= ([2(sqrtx-3)]/[(sqrtx+3)(sqrtx-3)]-(sqrtx-5)/[(sqrtx+3)(sqrtx-3)])*(sqrtx-3)/(sqrtx-1)` $\\$ `= (2sqrtx-6-sqrtx+5)/[(sqrtx+3)(sqrtx-3)]*(sqrtx-3)/(sqrtx-1)` $\\$ `= (sqrtx - 1)/[(sqrtx+3)(sqrtx-3)]*(sqrtx-3)/(sqrtx-1)=1/(sqrtx+3)`
`3)` `A-B = (2sqrtx+1)/(sqrtx+3)-1/(sqrtx+3)=(2sqrtx+1-1)/(sqrtx+3)=(2sqrtx)/(sqrtx+3)=2-6/(sqrtx+3)`
`6 vdotssqrtx+3=>sqrtx+3inƯ(6)={pm1;pm2;pm3;pm6}`
+) `sqrtx + 3=1=>sqrtx=-2=>x = emptyset`
+) `sqrtx + 3 = 2=>sqrtx= -5 => x = emptyset`
+) `sqrtx+3=-1=>sqrtx=-4=> x = emptyset`
+) `sqrtx + 3 = -2=>sqrtx=-5=> x = emptyset`
+) `sqrtx + 3 = 3=>sqrtx=0=>x=0`
+) `sqrtx + 3 = -3=>sqrtx=-6=> x = emptyset`
+) `sqrtx + 3 = 6=>sqrtx=3=>x=9`
+) `sqrtx + 3 = -6=>sqrtx=-9=>x=emptyset`
Vậy `x in {0;9}` để `A-B` có giá trị là số tự nhiên
