$$\begin{gathered} y=\frac{1}{x^2-1}+\sqrt{x^2-8x} \\ đkxđ: \left\{\begin{array}{l}{x}^2-1\ne 0\\ x^2-8x\ge 0\end{array}<=>\left\{\begin{array}{l}x\ne \pm 1\\ x.\left(x-8\right)\ge 0 \left(1\right)\end{array}\right.\right. \\ Giải \left(1\right): x.\left(x-8\right)=0<=>x=0; x=8 \\ x 0 8 \\ x - 0 + | + \\ x-8 - 0 - 0 + \\ f\left(x\right) + 0 - 0 + \\ =>\left(1\right)<=>x\le 0 hoặc x\ge 8 \\ cùng x\ne \pm 1 đuợc: \left\{\begin{array}{l}x\le 0\\ x\ne -1\end{array}hoặc x\ge 8\right. \\ Vậy: \left\{\begin{array}{l}x\le 0\\ x\ne -1\end{array}hoặc x\ge 8\right. \end{gathered}$$.

y=1x2−1+√x2−8xđkxđ: {x2−1≠0x2−8x≥0<=>{x≠±1x.(x−8)≥0 (1)Giải (1): x.(x−8)=0<=>x=0; x=8x                                0                    8 x                     −       0         +        ∣∣∣       +x−8                −      0         −        0      +f(x)                  +      0         −        0       +=>(1)<=>x≤0 hoặc x≥8cùng x≠±1 đuợc: {x≤0x≠−1hoặc x≥8Vậy: {x≤0x≠−1hoặc x≥8y=1x2-1+x2-8xđkxđ: x2-1≠0x2-8x≥0<=>x≠±1x.x-8≥0 1Giải 1: x.x-8=0<=>x=0; x=8x                                0                    8 x                     -       0         +        |       +x-8                -      0         -        0      +fx                  +      0         -        0       +=>1<=>x≤0 hoặc x≥8cùng x≠±1 đuợc: x≤0x≠-1hoặc x≥8Vậy: x≤0x≠-1hoặc x≥8.