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a fost răspuns

Să se rezolve inecuațiile:
a) [tex] x^{2} [/tex]-3x+4≥3x+2;
b) 2[tex] x^{2}[/tex]-2x<-[tex] \frac{1}{2} [/tex];
c) -[tex] x^{2} [/tex]-3x+5<[tex] x^{2} [/tex]-1;
d) [tex] x^{2} [/tex]+x+7≤0;
e) -[tex] x^{2} [/tex]+2x>x+[tex] \frac{1}{4} [/tex].

Răspuns :

Goodman10
1) x²-3x+4≥3x+2
x²-3x+4-3x-2≥0
x²-6x+2≥0
x²-6x+2=0
x=3+√7
x=3-√7
1(x-(3+√7))·(x-(3-√7))≥0
(x-3-√7)·(x-3+√7)≥0
[tex] \left \{ {{x-3- \sqrt{7} } \geq 0\atop {x-3+ \sqrt{7} \geq 0 }} \right. [/tex]

[tex] \left \{ {{x-3- \sqrt{7} \leq 0 } \atop {x-3+ \sqrt{7} \leq 0 }} \right. [/tex]

[tex] \left \{ {{x \geq 3+ \sqrt{7} } \atop {x=3- \sqrt{7} }} \right. [/tex]

x∈[3+√7,+∞}
x∈{-∞,3-√7]
x∈{-∞,3-√7]∪[3+√7,+∞}

3) x²+x+7≤0
x²+x+7=0
x∉R
x²+x+7≤0,a=1
x∈∅
succes!!!!



Vezi imaginea Goodman10
Vezi imaginea Goodman10

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