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Question Number 168515 by cortano1 last updated on 12/Apr/22

$$\mid \frac{x^2-9}{3}\mid +\frac{x-3}{3}>9$$ $$x=?$$

Commented bybenhamimed last updated on 12/Apr/22

$$\{\begin{array}{l}{x}^2+x-39>0six\in ]-\mathrm{\infty };-3]\cup [3;+\mathrm{\infty }[\\ -x^2+x-21>0six\in [-3;3]\end{array}$$ $$S=]-\mathrm{\infty };\frac{-1-\sqrt{157}}{2}[\cup ]\frac{-1+\sqrt{157}}{2};+\mathrm{\infty }[$$ $$$$

Answered by Mathspace last updated on 12/Apr/22

$$\Rightarrow \frac{\mid x^2-9\mid }{3}+\frac{x-3}{3}>9\Rightarrow$$ $$\mid x^2-9\mid +x-3>27?\Rightarrow$$ $$\mid x^2-9\mid +x-3-27>0\Rightarrow$$ $$\mid x^2-9\mid +x-30>0$$ $$f\left(x\right)=\mid x^2-9\mid +x-30$$ $$x-\mathrm{\infty }-33+\mathrm{\infty }$$ $$\mid x^2-9x^2-99-x^2{x}^2-9$$ $$x⩽-3\Rightarrow f\left(x\right)=x^2-9+x-30$$ $$=x^2+x-39$$ $$x\in [-3,3]\Rightarrow f\left(x\right)=9-x^2+x-30$$ $$=-x^2+x-21$$ $$x\in [3,+\mathrm{\infty }[\Rightarrow f\left(x\right)=x^2-9+x-30$$ $$=x^2+x-39$$ $$case1x⩽-3$$ $$f>0\Rightarrow x^2+x-39>0$$ $$\mathrm{\Delta }=1-4.(-39)=1+4.39$$ $$=1+156=157$$ $$x_1=\frac{-1+\sqrt{157}}{2}$$ $$x_2=\frac{-1-\sqrt{157}}{2}$$ $$....$$ $$$$

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