Решите неравенство: |х^2-10|>9x

Question 1

Question 2

A\; \; \; \to \; \; \; f(x)>A\; \; ili\; \; f(x)<-A\\\\\\|x^2-10|>9x\\\\1)\; x^2-10>9x\\\\x^2-9x-10>0,\; \; x_1=-1,x_2=10\\\\(x-10)(x+1)>0,\; \; +++(-1)---(10)+++\\\\x\in (-\infty,-1)U(10,+\infty)\\\\2)x^2-10<-9x\\\\x^2+9x-10<0,x_1=1,x_2=-10" alt="|f(x)|>A\; \; \; \to \; \; \; f(x)>A\; \; ili\; \; f(x)<-A\\\\\\|x^2-10|>9x\\\\1)\; x^2-10>9x\\\\x^2-9x-10>0,\; \; x_1=-1,x_2=10\\\\(x-10)(x+1)>0,\; \; +++(-1)---(10)+++\\\\x\in (-\infty,-1)U(10,+\infty)\\\\2)x^2-10<-9x\\\\x^2+9x-10<0,x_1=1,x_2=-10" align="absmiddle" class="latex-formula">

$(x-1)(x+10)<0,\; \; \ +++(-10)---(1)+++\\\\x\in (-10,1)$

3) Объудиняем множества, получим

$x\in (-\infty,1)U(10,+\infty)$

NNNLLL54_zn · Answer 1 · 2018-03-21T05:15:28+0000

A\; \; \; \to \; \; \; f(x)>A\; \; ili\; \; f(x)<-A\\\\\\|x^2-10|>9x\\\\1)\; x^2-10>9x\\\\x^2-9x-10>0,\; \; x_1=-1,x_2=10\\\\(x-10)(x+1)>0,\; \; +++(-1)---(10)+++\\\\x\in (-\infty,-1)U(10,+\infty)\\\\2)x^2-10<-9x\\\\x^2+9x-10<0,x_1=1,x_2=-10" alt="|f(x)|>A\; \; \; \to \; \; \; f(x)>A\; \; ili\; \; f(x)<-A\\\\\\|x^2-10|>9x\\\\1)\; x^2-10>9x\\\\x^2-9x-10>0,\; \; x_1=-1,x_2=10\\\\(x-10)(x+1)>0,\; \; +++(-1)---(10)+++\\\\x\in (-\infty,-1)U(10,+\infty)\\\\2)x^2-10<-9x\\\\x^2+9x-10<0,x_1=1,x_2=-10" align="absmiddle" class="latex-formula">

$(x-1)(x+10)<0,\; \; \ +++(-10)---(1)+++\\\\x\in (-10,1)$

3) Объудиняем множества, получим

$x\in (-\infty,1)U(10,+\infty)$