Răspuns:
a. [tex]4x{2} -11x-3\\[/tex]≤0
[tex]4x^{2} +x-12x-3\\[/tex]≤0
x(4x+1)-3(4x+1)≤0
(4x+1)(x-3)≤0
[tex]\left \{ {{4x+1\leq0 } \atop {x-3\geq 0}} \right.[/tex]
x∈[-[tex]\frac{1}{4}[/tex],3]
b. x (x-5)<2([tex]x^{2}[/tex]-4x+3)
[tex]x^{2} -5x\leq 2x^{2} -8x+6[/tex] (mutăm expresia în stânga)
[tex]x^{2} -5x-2x^{2} +8x-6<0[/tex](calculăm termenii asemenea)
[tex]-x^{2} +3x-6<0\\-x^{2} +3x-6=0\\[/tex]
⇒X∉R
c.
[tex]((2x-1)^{2} -x(x+1)\leq (x-3)^{2} -7\\4x^{2} -4x+1-x(x+1)\leq (x-3)^{2}-7\\3x^{2} -4x+1 -x^{2} -x\leq x^{2} -6x+9-7\\\\3x^{2} -5x+1\leq x^{2} -6x+2\\\\[/tex]
[tex]3x^{2} -5x+1-x^{2} +6x-2\leq 0\\2x^{2} +x-1\leq 0\\2x^{2} +2x-x-1\leq 0\\2x(x+1)-(x+1)\leq 0\\(x+1)(2x-1)\leq 0\\\left \{ {{x+1\geq 0 } \atop {2x-1\leq 0}} \right. \\[/tex]
[tex]\left \{ {{x\leq -1} \atop {x\geq1/2 }} \right.[/tex]
x∈[-1,[tex]\frac{1}{2}[/tex]]
Explicație pas cu pas:
