Đáp án:
Giải thích các bước giải:
Câu 12:
`y=\frac{x²+x+1}{x+1}`
`=> y' =\frac{(x²+x+1)'(x+1)-(x²+x+1)(x+1)'}{(x+1)²}`
`=>y' =\frac{(2x+1)(x+1)-(x²+x+1).1}{(x+1)²}`
`=> y' =\frac{2x²+2x+x+1-x²-x-1}{(x+1)²}`
`=> y' =\frac{x²+2x}{(x+1)²}`
Câu 13:
`y=\frac{x²-x+1}{x²+x+1}`
`=> y'=\frac{(x²-x+1)'(x²+x+1)-(x²-x+1)(x²+x+1)'}{(x²+x+1)²}`
`=> y' =\frac{(2x-1)(x²+x+1)-(x²-x+1)(2x+1)}{(x²+x+1)²}`
`=> y'=\frac{2x³ +2x² +2x -x²-x-1 -2x³+2x²-2x-x²+x-1}{(x²+x+1)²}`
`=>y'=\frac{2x²-2}{(x²+x+1)²}`
