Đáp án:
`1)` `3/{3x+5}`
`2)` `{2x+3}/{x^2-x+1}`
Giải thích các bước giải:
`1)` `3x^2+14x+15=3x^2+9x+5x+15`
`=3x(x+3)+5(x+3)=(x+3)(3x+5)`
Ta có:
`\qquad 1/{x+3}+4/{3x^2+14x+15}\quad (x\ne -3;x\ne -3/5)`
`=1/{x+3} +4/{(x+3)(3x+5)}`
`={3x+5}/{(x+3)(3x+5)}+4/{(x+3)(3x+5)}`
`={3x+5+4}/{(x+3)(3x+5)}={3(x+3)}/{(x+3)(3x+5)}`
`=3/{3x+5}`
Vậy: `1/{x+3}+4/{3x^2+14x+15}`
`=3/{3x+5}\quad (x\ne -3;x\ne -3/5)`
$\\$
`2)` `{6x}/{x^3+1}+1/{x^2-x+1}+2/{x+1}\ quad (x\ne -1)`
`={6x}/{(x+1)(x^2-x+1)}+{x+1}/{(x+1)(x^2-x+1)}+{2(x^2-x+1)}/{(x+1)(x^2-x+1)}`
`={6x+x+1+2x^2-2x+2}/{(x+1)(x^2-x+1)}`
`={2x^2+5x+3}/{(x+1)(x^2-x+1)}`
`={2x^2+2x+3x+3}/{(x+1)(x^2-x+1)}`
`={2x(x+1)+3(x+1)}/{(x+1)(x^2-x+1)}`
`={(x+1)(2x+3)}/{(x+1)(x^2-x+1)}={2x+3}/{x^2-x+1}`
Vậy: `{6x}/{x^3+1}+1/{x^2-x+1}+2/{x+1}`
`={2x+3}/{x^2-x+1} \ quad (x\ne -1)`
