Question

As vrea și eu rezolvarea la toate exercițiile

Answer 1

$\displaystyle 1.~x*y=x+y-\frac{xy}{4} \\ \\ a)~6*2=5 \\ \\ 6*2=6+2-\frac{6 \cdot 2}{4} =8-\frac{12}{4} =8-3=\mathbf5\\ \\ b)~x*(4x)=6\Rightarrow x+4x-\frac{x \cdot 4x}{4} =6\Rightarrow 5x-\frac{4x^2}{4}=6\Rightarrow 5x-x^2=6\Rightarrow\\\\ \Rightarrow-x^2+5x-6=0 \Rightarrow x^2-5x+6=0\Rightarrow x^2-3x-2x+6=0 \Righrarrow\\\\\Rightarrow x(x-3)-2(x-3)=0\Rightarrow (x-2)(x-3)=0\\\\x-2=0\Rightarrow x=0+2\Rightarrow \mathbf{x=2}\\\\x-3=0\Rightarrow x=0+3\Rightarrow \mathbf{x=3}~~~~~~~~~~~~~~~~~~~~~~~~S=\{2;3\}$

$\displaystyle 2.~x \circ y=xy-9(x+y)+90\\ \\ a)~10 \circ 8=8\\ \\ 10 \circ 8=10 \cdot 8-9(10+8)+90=80-9 \cdot 18+90=80-162+90=\\ \\ =-82+90=\mathbf8\\ \\ b)~x \circ y=(x-9)(y-9)+9\\\\x\circ y=xy-9(x+y)+90=xy-9x-9y+90=\\\\=xy-9x-9y+81+9=x(y-9)-9(y-9)+9=\\\\=\mathbf{(x-9)(y-9)+9}$

$\displaystyle c)~n\circ n\leq10\Rightarrow n\cdot n-9(n+n)+90\leq 10\Rightarrow n^2-9 \cdot2n+90-10\leq0\Rightarrow\\\\ \Rightarrow n^2-18n+80\leq 0\Rightarrow n^2-8n-10n+80\leq 0\Rightarrow \\ \\\Rightarrow n(n-8)-10(n-8)\leq 0\Rightarrow(n-8)(n-10)\leq0 \\ \\ (n-8)(n-10)=0 ~~~~~~~~~~~~~~~~~~~~~~n-8=0 \Rightarrow n=0+8\Rightarrow n=8\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~n-10=0\Rightarrow n=0+10\Rightarrow n=10$

$\displaystyle\left\begin{array}{ccc}n\\n-8\\n-10\\(n-8)(n-10)\end{array}\right|\left\begin{array}{ccc}-\infty~~~~~~~~~~~~~~~8~~~~~~~~~~~~~~~~~~~~~~~10~~~~~~~~~~~~~~~~~~~~+\infty\\------0+++++++++++++++++\\---------------0++++++++\\++++++0--------~0++++++++\end{array}\right$

$\displaystyle \mathbf{n \in [8;10]}$

$\displaystyle3.~x*y=xy-(x+y)+2\\ \\ a)~2*2=2\\ \\ 2*2=2 \cdot 2-(2+2)+2=4-4+2=0+2=\mathbf2\\ \\ b)~x*y=(x-1)(y-1)+1\\ \\ x*y=xy-(x+y)+2=xy-x-y+2=xy-x-y+1+1=\\ \\ =x(y-1)-(y-1)+1=\mathbf{(x-1)(y-1)+1}$

$\displaystyle 4.~x\circ y=5xy+15(x+y)+42\\ \\ a)~(-2)\circ(-2)=2\\ \\ (-2)\circ(-2)=5 \cdot (-2) \cdot (-2)+15(-2-2)+42=20+15 \cdot (-4)+42=\\ \\ =20-60+42=-40+42=\mathbf2\\ \\ b)~x\circ y=5(x+3)(y+3)-3 \\ \\ x \circ y=5xy+15(x+y)+42=5xy+15x+15y+42=\\ \\ =5xy+15x+15y+45-3=5x(y+3)+15(y+3)-3=\\ \\ =(5x+15)(y+3)-3=\mathbf{5(x+3)(y+3)-3}$

$\displaystyle 5.~x*y=xy-2(x+y)+6\\ \\ a)~x*y=(x-2)(y-2)+2 \\ \\ x*y=xy-2(x+y)+6=xy-2x-2y+6=xy-2x-2y+4+2=\\ \\ =x(y-2)-2(y-2)+2=\mathbf{(x-2)(y-2)+2}\\ \\ b)~x\circ 3=2018 \Rightarrow x \cdot 3-2(x+3)+6=2018 \Rightarrow 3x-2x-6+6=2018\Rightarrow \\ \\ \Rightarrow \mathbf{x=2018}$

$\displaystyle 6.~x\circ y=3xy+3x+3y+2\\ \\ a)~x \circ y=3(x+1)(y+1)-1\\ \\ x \circ y=3xy+3x+3y+2=3xy+3x+3y+3-1=\\ \\ =3x(y+1)+3(y+1)-1=(3x+3)(y+1)-1=\mathbf{3(x+1)(y+1)-1}\\ \\ b)~x \circ \left(-\frac{2}{3} \right)=x \\ \\ x \circ \left(-\frac{2}{3} \right)=3 \cdot x \cdot \left(-\frac{2}{3} \right)+3 \cdot x+3 \cdot \left(-\frac{2}{3} \right)+2=-2x+3x-2+2=\mathbf x$

$\displaystyle c)~n\circ (n-1)<17\Rightarrow 3n(n-1)+3n+3(n-1)+2<17 \Rightarrow \\ \\ \Rightarrow 3n^2-3n+3n+3n-3+2-17<0 \Rightarrow 3n^2+3n-18<0\Big|:3 \Rightarrow \\ \\ \Rightarrow n^2+n-6<0 \Rightarrow n^2+3n-2n-6<0\Rightarrow n(n+3)-2(n+3)<0 \Rightarrow \\ \\ \Rightarrow (n-2)(n+3)<0\\ \\ (n-2)(n+3)=0~~~~~~~~~~~~~~~~~~~~~~~~~~n-2=0\Rightarrow n=0+2\Rightarrow n=2\\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~n+3=0\Rightarrow n=0-3 \Rightarrow n=-3$

$\displaystyle\left\begin{array}{ccc}n\\n-2\\n+3\\(n-2)(n+3)\end{array}\right|\left\begin{array}{ccc}-\infty~~~~~~~~~~~~~~~-3~~~~~~~~~~~~~~~~~~~~~~2~~~~~~~~~~~~~~~~~~~~+\infty\\----------------0+++++++++\\--------0+++++++++++++++++\\++++++++0-------~0+++++++++\end{array}\right$

$\displaystyle \mathbf{n\in (-3;2)}$