Question

Va rog frumos problemele care sunt incercuite.2 c,4a si 5 tot.Multumesc!

Answer 1

$3-2x \leq x-2 \Leftrightarrow3+2 \leq x+2x \Leftrightarrow 5 \leq 3x\Rightarrowx \geq \frac{5}{3} . \\ Cazuri~favorabile:~2~si~3~(in~total~sunt~doua). \\ \\ P= \frac{Nr.caz.fav.}{Nr.caz.pos.}= \frac{2}{5} (=40\%).$
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$\frac{x+1}{ \sqrt{3}+2 }+ \sqrt{3}=2-x \\ \frac{x+1}{ \sqrt{3}+2 }+ \sqrt{3} -2=-x \\ \frac{(x+1)( \sqrt{3}-2) }{3-4}+ \sqrt{3}-2=-x \\ -(x+1)( \sqrt{3}-2)+ \sqrt{3}-2=-x~~~~~~~~~~|*(-1) \\ (x+1)( \sqrt{3}-2)-( \sqrt{3}-2)=x \\ (\sqrt{3}-2)(x+1-1)=x \\ ( \sqrt{3}-2)*x=x \Rightarrow \boxed{x=0}.$
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$a)~E(-1;1)=(-1-2)(-1-4)+(1+1)(1-5)-5= \\ =-3*(-5)+2*(-4)-5= \\ =15-8-5= \\ =2.$

$b)~Am~avut~si~eu~acelasi~manual.~La~acest~punct~s-a~strecurat~o \\ greseala:~dupa~" \geq "~urma~"-10",~nu~"b-10". \\ \\ (a-2)(a-4)+(b+1)(b-5)= \\ =[(a-3)+1][(a-3)-1]+[(b-2)+3][(b-2)-3]= \\ =(a-3) ^{2}-1 +( b-2)^{2} -9= \\ =(a-3) ^{2} +(b-2) ^{2} -10 \geq 0+0-10=-10.$