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Tìm các điểm hàm số đi qua với mọi my = x 3 + m x 2 − m − 1 ↔ m ( x 2 − 1 ) + ( x 3 − y − 1 ) = 0 m ọ i m ↔ { x 2 − 1 = 0 x 3 − y − 1 = 0 ↔ { x = 1 → y = 0 x = − 1 → y = − 2 y ′ = 3 x 2 + 2 m x + x = 1 , y = 0 → y ′ = 3 + 2 m → P T T T : y = ( 3 + 2 m ) ( x − 1 ) + x = − 1 , y = − 2 → y ′ = 3 − 2 m → P T T T : y = ( 3 − 2 m ) ( x + 1 ) − 2 y=x^3+mx^2-m-1 \leftrightarrow m(x^2-1)+(x^3-y-1)=0 \ mọi \ m \leftrightarrow \begin{cases} x^2-1=0 \\ x^3-y-1=0 \end{cases} \leftrightarrow \begin{cases} x=1 \rightarrow y=0 \\ x=-1 \rightarrow y=-2 \end{cases} \\ y'=3x^2+2mx \\ + x=1, y=0 \rightarrow y'=3+2m \rightarrow PTTT: \ y=(3+2m)(x-1) \\ +x=-1, \ y=-2 \\ \rightarrow y'=3-2m \rightarrow PTTT: \ y=(3-2m)(x+1)-2 y = x 3 + m x 2 − m − 1 ↔ m ( x 2 − 1 ) + ( x 3 − y − 1 ) = 0 m ọ i m ↔ { x 2 − 1 = 0 x 3 − y − 1 = 0 ↔ { x = 1 → y = 0 x = − 1 → y = − 2 y ′ = 3 x 2 + 2 m x + x = 1 , y = 0 → y ′ = 3 + 2 m → P T T T : y = ( 3 + 2 m ) ( x − 1 ) + x = − 1 , y = − 2 → y ′ = 3 − 2 m → P T T T : y = ( 3 − 2 m ) ( x + 1 ) − 2
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