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Radik2015
a fost răspuns

Aflatil pe x 1)2(x-1)≥3(x+2) 2)4(3-x)+x<3(x+2) 3)3(x+4)+(x+1) (x-1)≤
≤ (x-2) (x+3) 2(3-x)-1 4)2x(x-1)+(x-3) (x+4)>x(3x+2)-5(1-x)+9 5)(2x+1)²+5(x-3)<4(x-2)²-10(x+2) Va roog

Răspuns :

Nitugheorghe57
1)2(x-1)≥3(x-2)⇒2x-2≥3x-6⇒2x-3x≥2-6⇒-x≥-4⇒x≥4
b)4(3-x)+x≤3(x+2)⇒12-4x+x≤3x+6⇒12-3x≤3x+6⇒-3x-3x≤-12+6⇒-6x≤-6⇒
x≤-6/-6⇒x≤1
sucees

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