Question

Va rog dau coroana , exercitiul 2 urgent
@andyilye​

Answer 1

$II.~~x \circ y=2xy-6x-6y+21\\\\ a)~x\circ y=2(x-3)(y-3)+3\\\\ x\circ y=2xy-6x-6y+21=2xy-6x-6y+18+3=\\\\=2x(y-3)-6(y-3)+3=(2x-6)(y-3)+3=2(x-3)(y-3)+3$

$\displaystyle b)~x \circ x=11;~5^x \circ 5^x=11\\\\ x\circ x=11 \Rightarrow 2(x-3)(x-3)+3=11 \Rightarrow 2(x-3)^2+3=11\Rightarrow \\\\\Rightarrow 2\Big(x^2-2 \cdot x \cdot 3+3^2\Big)+3=11\Rightarrow 2\Big(x^2-6x+9\Big)+3=11\Rightarrow \\\\\Rightarrow 2x^2-12x+18+3-11=0\Rightarrow 2x^2-12x+10=0\\\\ \Delta=(-12)^2-4 \cdot 2 \cdot 10=144-80=64 > 0\\\\ x_1=\frac{-(-12)-\sqrt{64} }{2 \cdot 2} =\frac{12-8}{4}=\frac{4}{4} =1\\\\ x_2=\frac{-(-12)+\sqrt{64} }{2 \cdot 2}=\frac{12+8}{4} =\frac{20}{4} =5$

$\displaystyle 5^x \circ 5^x=11 \Rightarrow 2\Big(5^x-3\Big)\Big(5^x-3\Big)+3=11 \Rightarrow 2\Big(5^x-3\Big)^2+3=11\Rightarrow \\\\ \Rightarrow 2\Big(\Big(5^x\Big)^2-2 \cdot 5^x \cdot 3+3^2\Big)+3=11\Rightarrow 2\Big(5^{2x}-6 \cdot 5^x+9\Big)+3=11\Rightarrow \\\\\Rightarrow 2 \cdot 5^{2x}-12 \cdot 5^x+18+3-11=0\Rightarrow 2 \cdot 5^{2x}-12 \cdot 5^x+10=0\\\\ 5^x=t\\\\ 2t^2-12t+10=0\\\\ \Delta=(-12)^2-4 \cdot 2 \cdot 10=144-80=64 > 0$

$\displaystyle t_1=\frac{-(-12)-\sqrt{64} }{2 \cdot 2} =\frac{12-8}{4} =\frac{4}{4} =1\\\\ t_2=\frac{-(-12)+\sqrt{64} }{2 \cdot 2} =\frac{12+8}{4} =\frac{20}{4} =5$

$5^x=t_1 \Rightarrow 5^x=1 \Rightarrow x=0\\\\ 5^x=t_2 \Rightarrow 5^x=5 \Rightarrow x=1$

$\displaystyle c)~x \circ e = e \circ x = x \\\\ x\circ e=x \Rightarrow 2(x-3)(e-3)+3=x \Rightarrow 2(x-3)(e-3)=x-3 \Rightarrow \\\\ \Rightarrow 2(e-3)=\frac{x-3}{x-3} \Rightarrow 2e-6=1\Rightarrow 2e=1+6 \Rightarrow 2e=7 \Rightarrow e=\frac{7}{2}$

$\displaystyle e \circ x=x \Rightarrow 2(e-3)(x-3)+3=x \Rightarrow 2(e-3)(x-3)=x-3 \Rightarrow \\ \\ \Rightarrow 2(e-3)=\frac{x-3}{x-3} \Rightarrow 2e-6=1 \Rightarrow 2e=1+6 \Rightarrow 2e=7 \Rightarrow e=\frac{7}{2}$

$\displaystyle x \circ x'=x' \cdot x=e\\\\ 2(x-3)(x'-3)+3=\frac{7}{2} \Rightarrow 2(x-3)(x'-3)=\frac{7}{2} -3\Rightarrow \\\\ \Rightarrow 2(x-3)(x'-3)=\frac{7-6}{2} \Rightarrow 2(x-3)(x'-3)=\frac{1}{2} \Rightarrow \\\\\Rightarrow (x-3)(x'-3)=\frac{\cfrac{1}{2} }{2} \Rightarrow (x-3)(x'-3)=\frac{1}{2} \cdot \frac{1}{2} \Rightarrow (x-3)(x'-3)=\frac{1}{4}\Rightarrow \\\\ \Rightarrow x'-3=\frac{\cfrac{1}{4} }{x-3} \Rightarrow x'-3=\frac{1}{4(x-3)} \Rightarrow x'=\frac{1}{4(x-3)} +3,~x\not =3$

$d)~1 \circ (2 \circ 3)=(1 \circ 2) \circ 3\\\\ 1 \circ(2\circ3)=1\circ(2(2-3)(3-3)+3)=1\circ(2((-1)\cdot0)+3)=\\\\ =1\circ (2 \cdot 0+3)=1 \circ 3=2(1-3)(3-3)+3=2((-2) \cdot 0)+3=\\\\ =2 \cdot 0+3=0+3=3\\\\ (1\circ2)\circ3=(2(1-3)(2-3)+3)\circ 3=(2((-2)\cdot(-1))+3) \circ 3=\\\\ =(2 \cdot 2+3) \circ 3=(4+3) \circ 3=7 \circ 3=2(7-3)(3-3)+3=\\\\=2(4 \cdot 0)+3=2 \cdot 0+3=0+3=3$

$\left\begin{array}{ccc}1\circ(2\circ3)=3\\(1\circ2)\circ3=3\end{array}\right\Bigg| \Rightarrow 1 \circ (2\circ3)=(1\circ2)\circ3=3$

Answer 2

Explicație pas cu pas:

a)

$x \circ y = 2xy - 6x - 6y + 21 = 2xy - 6x - 6y + 18 + 3 = 2x(y - 3) - 6(y - 3) + 3 = (2x - 6)(y - 3) + 3 = \bf 2(x - 3)(y - 3) + 3$

b)

▪︎

$x \circ x = 11$

$2(x - 3)^{2} + 3 = 11$

$(x - 3)^{2} = 4 \iff |x - 3| = 2$

$x - 3 = -2 \implies x = 1$

$x - 3 = 2 \implies x = 5$

▪︎

$5^{x} \circ 5^{x} = 11$

$2(5^{x} - 3)^{2} + 3 = 11$

$(5^{x} - 3)^{2} = 4 \iff |5^{x} - 3| = 2$

$5^{x} - 3 = -2 \iff 5^{x} = 1 \\ 5^{x} = 5^{0} \implies x = 0$

$5^{x} - 3 = 2 \iff 5^{x} = 5 \\ 5^{x} = 5^{1} \implies x = 1$

c)

element neutru:

$x \circ e = e \circ x = x$

$x \circ e = x$

$2ex - 6x - 6e + 21 = x$

$x(2e - 7) - 3(2e - 7) = 0$

$(x - 3)(2e - 7) = 0$

$2e - 7 = 0 \implies e = \dfrac{7}{2}$

$e \circ x = x$

$2ex - 6e - 6x + 21 = x$

$2e(x - 3) - 7(x - 3) = 0$

$(x - 3)(2e - 7) = 0$

$2e - 7 = 0 \implies e = \dfrac{7}{2}$

element simetrizabil:

$x \circ x^{\prime} = x^{\prime} \circ x = e$

$x \circ x^{\prime} = e$

$2xx^{\prime} - 6x - 6x^{\prime} + 21 = \dfrac{7}{2} \\ 2x^{\prime}(x - 3) = 6x - \dfrac{35}{2} \\ x^{\prime} = \dfrac{12x - 35}{4(x - 3)}\ , \ \forall \ x \neq 3$

$x^{\prime} \circ x = e$

$2xx^{\prime} - 6x^{\prime} - 6x + 21 = \dfrac{7}{2} \\ 2x^{\prime}(x - 3) = 6x - \dfrac{35}{2} \\ x^{\prime} = \dfrac{12x - 35}{4(x - 3)}\ , \ \forall \ x \neq 3$

d)

$1 \circ (2 \circ 3) = 1 \circ (2(2 - 3)(3 - 3) + 3) = 1 \circ (0 + 3) = 1 \circ 3 = 2(1 - 3)(3 - 3) + 3 = 0 + 3 = 3$

$(1 \circ 2) \circ 3 = (2(1 - 3)(2 - 3) + 3) \circ 3 = (4 + 3) \circ 3 = 7 \circ 3 = 2(7 - 3)(3 - 3) + 3 = 0 + 3 = 3$

$\implies 1 \circ (2 \circ 3) = (1 \circ 2) \circ 3$

q.e.d.