Question

va rog e urgent dau coroana dau
exercițiul 5 din imagine
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Answer 1

$\displaystyle\bf\\x^2+y^2-2(2x-3y)+9=0 \Leftrightarrow x^{2}+y^{2}-4x+6y+9=0 \Leftrightarrow\\adunam~in~ambii~membri \implies x^2+y^2-4x+6y+9+4=4 \Leftrightarrow\\x^2-4x+4+y^2+6y+9=4 \Leftrightarrow (x-2)^2+(y+3)^2=4.\\(x-2)^2\leq 4|\sqrt()~ \Leftrightarrow |x-2|\leq 2 \implies x \in [0,4] .\\(y+3)^3\leq 4|\sqrt()~ \Leftrightarrow |y+3|\leq 2 \implies y \in [-5,-1] .\\asadar~cum~x~poate~lua~doar~valori~pozitive~iar~y~doar~valori~negative~\\ \implies x > y.$

$\displaystyle\bf\\E_1(x)=4x^3+8x^2-x-2=4x^2(x+2)-x-2=4x^2(x+2)-(x+2)=\\(x+2)((2x)^2-1^2)=(x+2)(2x-1)(2x+1).\\E_2(x)=2x^3+x^2-8x-4=x^2(2x+1)-4(2x+1)=(2x+1)(x^2-2^2)=\\(2x+1)(x-2)(x+2).\\E_1(x)-E_2(x)=0 \Leftrightarrow (2x+1)(2x-1)(x+2)-(2x+1)(x-2)(x+2)=0\Leftrightarrow(2x+1)(x+2)(2x-1-x+2) = 0 \Leftrightarrow (2x+1)(x+2)(x+1)=0.\\este~suficient~ca~una~dintre~cele~trei~paranteze~sa~fie~0~, deci~vom~avea~trei~solutii.\\1.~2x+1=0 \implies x = -\frac{1}{2} .\\2.~x+2=0 \implies x = -2.$$\displaystyle\bf\\3.~x+1=0 \implies x=-1.$