Question

Va rog dau coroana ex 8 ​

Answer 1

$x \bot y=(x-3)(y-3)+3\\\\ a)~(x+3)\bot \big(\frac{1}{x} +3\big)=4,~~(\forall)x\in\mathbb{R}^*\\\\(x+3)\bot \big(\frac{1}{x} +3\big)+3=(x+3-3)\big(\frac{1}{x} +3-3)+3=\not x \cdot \frac{1}{\not x} +3=1+3=4$

$\displaystyle b)~e=4\\\\ x\bot e=e\bot x=x\\\\ x\bot e=x\Rightarrow (x-3)(e-3)+3=x\Rightarrow (x-3)(e-3)=x-3\Rightarrow \\\\ \Rightarrow e-3=\frac{x-3}{x-3}\Rightarrow e-3=1\Rightarrow e=1+3 \Rightarrow e=4$

$\displaystyle e \bot x=x \Rightarrow (e-3)(x-3)+3=x\Rightarrow (e-3)(x-3)=x-3 \Rightarrow \\\\ \Rightarrow e-3=\frac{x-3}{x-3} \Rightarrow e-3=1 \Rightarrow e=1+3 \Rightarrow e=4$

$\displaystyle c)~x\bot x'=x' \bot x=e\\\\ x\bot x'=e \Rightarrow (x-3)(x'-3)+3=4 \Rightarrow (x-3)(x'-3)=4-3 \Rightarrow \\\\ \Rightarrow (x-3)(x'-3)=1 \Rightarrow x'-3=\frac{1}{x-3} \Rightarrow x'=\frac{1}{x-3} +3,~x\not=3$

$\displaystyle x'\bot x=e \Rightarrow (x'-3)(x-3)+3=4\Rightarrow (x'-3)(x-3)=4-3 \Rightarrow \\\\\Rightarrow (x'-3)(x-3)=1 \Rightarrow x'-3=\frac{1}{x-3}\Rightarrow x'=\frac{1}{x-3}+3,~x\not=3$