Giải thích các bước giải:
$\text{Ta có: ĐKXĐ:}$ `2x+1 # 0`
`<=> x # -1/2`
`=> D = R ``/`` {-1/2}`
`+)`` y^' = (sqrt(x^2 - 2x +2) / (2x +1))^'=(((2x-2)(2x +1))/(2sqrt(x^2 - 2x +2))-2sqrt(x^2 - 2x +2)) / (2x +1)^2`
`=((4x^2-2x-2)/(2sqrt(x^2 - 2x +2))-(4(x^2 - 2x +2))/(2sqrt(x^2 - 2x +2))) / (2x +1)^2=((4x^2-2x-2)/(2sqrt(x^2 - 2x +2))-(4x^2 - 8x +8)/(2sqrt(x^2 - 2x +2))) / (2x +1)^2`
`=((6x-10)/(2sqrt(x^2 - 2x +2))) / (2x +1)^2=(6x-10)/(2sqrt(x^2 - 2x +2)(2x +1)^2)`
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`Cho` `y^' = 0; => (6x-10)/(2sqrt(x^2 - 2x +2)(2x +1)^2) = 0`
`<=> x = 5/3; ĐK: x # -1/2`
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$\text{Ta có bảng:}$
