`a)``Q=(1/(x+1)+(6x+3)/(x^3+1)-2/(x^2-x+1)):(x+2)(ĐKXĐ:xne-1;xne-2)`
`Q=((x^2-x+1)/((x+1)(x^2-x+1))+(6x+3)/((x+1)(x^2-x+1))-(2(x+1))/((x+1)(x^2-x+1)))*1/(x+2)`
`Q=(x^2-x+1+6x+3-2x-2)/((x+1)(x^2-x+1))*1/(x+2)`
`Q=(x^2+3x+2)/((x+1)(x^2-x+1))*1/(x+2)`
`Q=((x+1)(x+2))/((x+1)(x^2-x+1))*1/(x+2)`
`Q=1/(x^2-x+1)`
`b)text(thay )x=1/3`
`toQ=1/((1/3)^2-1/3+1)=1/(7/9)=9/7`
`c)x^2-x+1`
`=(x^2-2*x*1/2+1/4)-1/4+1`
`=(x-1/2)^2+3/4>=3/4AAx`
`to1/((x-1/2)^2-1/4)<=1/(3/4)=-4AAx`
`to1/(x^2-x+1)<=4/3AAx`
`text(Dấu "=" xảy ra khi )`
`x-1/2=0`
`<=>x=1/2`
`text(Vậy )Q_max=4/3<=>x=1/2`
