Bài 1: `ĐKXĐ:x\ne+-3`
`a,` `P=({x+3}/{2x-6}-{x-3}/{2x+6}-{6}/{9-x^2}):{x+1}/{x-3}`
`=[{x+3}/{2(x-3)}-{x-3}/{2(x+3)}+{6}/{x^2-9}]:{x+1}/{x-3}`
`=[{x+3}/{2(x-3)}-{x-3}/{2(x+3)}+{6}/{(x+3)(x-3)}]:{x+1}/{x-3}`
`=[{(x+3)(x+3)}/{2(x-3)(x+3)}-{(x-3)(x-3)}/{2(x+3)(x-3)}+{12}/{2(x+3)(x-3)}]:{x+1}/{x-3}`
`={(x+3)(x+3)-(x-3)(x-3)+12}/{2(x-3)(x+3)}:{x+1}/{x-3}`
`={(x+3)^2-(x-3)^2+12}/{2(x-3)(x+3)}:{x+1}/{x-3}`
`={x^2+6x+9-(x^2-6x+9)+12}/{2(x-3)(x+3)}:{x+1}/{x-3}`
`={x^2+6x+9-x^2+6x-9+12}/{2(x-3)(x+3)}:{x+1}/{x-3}`
`={12x+12}/{2(x-3)(x+3)}:{x+1}/{x-3}`
`={12(x+1)}/{2(x-3)(x+3)}.{x-3}/{x+1}`
`={6}/{x+3}`
Vậy với `x\ne+-3` thì `P={6}/{x+3}`
`b,` `x=1/5` (Thỏa mãn ĐKXĐ)
Thay `x=1/5` vào `P` có:
`P={6}/{1/5+3}=6/(16/5)=6.5/16=30/16=15/8`
Vậy với `x=1/5` thì `P=15/8`
Bài 2: `ĐKXĐ:x\ne+-2`
`a,` `P=({x+2}/{2x-4}-{x-2}/{2x+4}-{4}/{4-x^2}):{x+1}/{x-2}`
`=[{x+2}/{2(x-2)}-{x-2}/{2(x+2)}+{4}/{x^2-4}]:{x+1}/{x-2}`
`=[{x+2}/{2(x-2)}-{x-2}/{2(x+2)}+{4}/{(x+2)(x-2)}]:{x+1}/{x-2}`
`=[{(x+2)(x+2)}/{2(x-2)(x+2)}-{(x-2)(x-2)}/{2(x+2)(x-2)}+{8}/{2(x+2)(x-2)}]:{x+1}/{x-2}`
`={(x+2)(x+2)-(x-2)(x-2)+8}/{2(x-2)(x+2)}:{x+1}/{x-2}`
`={(x+2)^2-(x-2)^2+8}/{2(x-2)(x+2)}:{x+1}/{x-2}`
`={x^2+4x+4-(x^2-4x+4)+8}/{2(x-2)(x+2)}:{x+1}/{x-2}`
`={x^2+4x+4-x^2+4x-4+8}/{2(x-2)(x+2)}:{x+1}/{x-2}`
`={8x+8}/{2(x-2)(x+2)}:{x+1}/{x-2}`
`={8(x+1)}/{2(x-2)(x+2)}.{x-2}/{x+1}`
`={4}/{x+2}`
Vậy với `x\ne+-3` thì `P={4}/{x+2}`
`b,` `x=-3/4` (Thỏa mãn ĐKXĐ)
Thay `x=-3/4` vào `P` có:
`P={4}/{-3/4+2}=4/(5/4)=4.4/5=16/5`
Vậy với `x=-3/4` thì `P=16/5`
