Question

DAU 35 DE PUNKTE AJUTATIMA VA ROGGG!!!!! Rezolva in R inecuatia :
a) 2(2x-3)+3(x-2)≥2;
b) 2x²-3x-2<0;
c) 9x²+12x+4>0;
d) -2x²+x-3≥0

Answer 1

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Answer 2

$a)~4x-6+3x-6 \geq 2\ \textless \ =\ \textgreater \ 7x-12 \geq 2=\ \textgreater \ 7x \geq 14=\ \textgreater \ x \geq 2.$
Solutie: x∈[2;+∞).

$b)~2 x^{2} -3x-2\ \textless \ 0\ \textless \ =\ \textgreater \ 2 x^{2} -4x+x-2\ \textless \ 0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ 2x(x-2)+(x-2)\ \textless \ 0\ \textless \ =\ \textgreater \ (x-2)(2x+1)\ \textless \ 0=\ \textgreater \ o~ \\ paranteza~este~mai~mica~decat~0,~iar~cealalta~mai ~mare~decat~0. \\ x-2\ \textless \ 0=\ \textgreater \ x\ \textless \ 2~si~2x+1\ \textgreater \ 0=\ \textgreater \ x\ \textgreater \ - \frac{1}{2} \\ x-2\ \textgreater \ 0=\ \textgreater \ x\ \textgreater \ 2~si~2x+1\ \textless \ 0=\ \textgreater \ x\ \textless \ -\frac{1}{2} ,~imposibil.$
Solutie: x∈$(- \frac{1}{2};2)$.

$c)~9x^{2} +12x+4\ \textgreater \ 0\ \textless \ =\ \textgreater \ (3x+2) ^{2} \ \textgreater \ 0. \\ (3x+2) ^{2} \geq0,pentru~orice~numar~real~x, \\ ~dar~nu~trebuie~sa~avem~egalitate=\ \textgreater \ 3x+2 \neq 0=\ \textgreater \ \\ x \neq -\frac{2}{3} .$
Solutie: x∈R-{$- \frac{2}{3}$}.

$d)~-2 x^{2} +x-3 \geq 0\ \textless \ =\ \textgreater \ 2 x^{2} -x+3 \leq 0\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (2 x^{2} -x+ \frac{1}{8}) + \frac{23}{8} \leq 0\ \textless \ =\ \textgreater \ ( \sqrt{2} x- \frac{ \sqrt{2} }{4} )+ \frac{23}{8} \leq 0,~imposibil.$
Inegalitatea nu are solutii reale.