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[tex] \frac{3}{7} x {}^{2} - \frac{1}{14} x - \frac{1}{2} \geqslant 0[/tex]
Решите неравенство:
[tex]1 \frac{1}{3} x {}^{2} - 2 \frac{5}{6} x + 1 \frac{1}{2} > 0[/tex]

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Universalka

Verified answer

$1)\frac{3}{7} x^{2}-\frac{1}{14}x-\frac{1}{2}\geq0\\\\\frac{3}{7} x^{2}*14-\frac{1}{14}x*14-\frac{1}{2}*14\geq0 \\\\6x^{2}-x-7\geq 0\\\\6x^{2}-x-7=0\\\\D=(-1)^{2}-4*6*(-7)=1+168=169=13^{2}\\\\x_{1}=\frac{1-13}{12}=-1\\\\x_{2}=\frac{1+13}{12}= 1\frac{1}{6}\\\\6x^{2} -x-7=6(x+1)(x-1\frac{1}{6})\\\\(x+1)(x-1\frac{1}{6})\geq 0$

+ - +

______[- 1]______[1 1/6]______

/////////////// //////////////////

x ∈ ( - ∞ ; - 1] ∪ [ 1 1/6 ; + ∞)

$2)1\frac{1}{3}x^{2}-2\frac{5}{6} x+1\frac{1}{2}>0\\\\\frac{4}{3}x^{2}*6-\frac{17}{6} x*6+\frac{3}{2}*6>0 \\\\8x^{2}-17x+9>0\\\\8x^{2}-17x+9=0\\\\D=(-17)^{2} -4*8*9=289-188=1\\\\x_{1}=\frac{17-1}{16}=1\\\\x_{2}=\frac{17+1}{16}=1,125\\\\8x^{2}-17x+9=8(x-1)(x-1,125)\\\\(x-1)(x-1,125)>0$

+ - +

______₀_______₀______

1 1,125

//////////// //////////////

x ∈ ( - ∞ ; 1) ∪ (1,125 ; + ∞)

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