[tex]\it a) \ 1*1 = \dfrac{3+1}2=\dfrac{4}{2}=2\\ \\ \\ b) (1*2)*3=\dfrac{6+1}{3}*3=\dfrac{7}{3}*3=\dfrac{21+1}{\dfrac{16}{3}}=\dfrac{22\cdot3}{16}=\dfrac{66}{16}\\ \\ \\ 1*(2*3)=1*\dfrac{18+1}{5}=1*\dfrac{19}{5}=\dfrac{\dfrac{3\cdot19}{5}+1}{1+\dfrac{19}{5}}=\dfrac{62}{24}[/tex]
[tex]\it \dfrac{^{3)}66}{16}-\dfrac{^{2)}62}{24}=\dfrac{198-124}{48}=\dfrac{124^{(4}}{48}=\dfrac{31}{12}[/tex]
[tex]\it x*x=2 \Rightarrow \dfrac{3x^2+1}{2x}=2 \Rightarrow 3x^2-4x+1=0 \Rightarrow 3x^2-x-3x+1=0 \Rightarrow \\ \\ \Rightarrow x(3x-1)-(3x-1)=0 \Rightarrow (3x-1)(x-1)=0 \Rightarrow x_1=\dfrac{1}{3},\ x_2=1[/tex]
