Решите неравенство а)7х+3>5(х-4)+1 б)2хХво второй степени +13х-7 >0 в)2(1-х)≥5х(3х+2)...

1.    $7x+3\ \textgreater \ 5(x-4)+1$
$7x+3\ \textgreater \ 5x-20+1$
$7x-5x\ \textgreater \ -19-3$
$2x\ \textgreater \ -22$
$x\ \textgreater \ -11$
2.    $2 x^{2} +13x-7\ \textgreater \ 0$
$D=169+56=225$
$x_1= \frac{-13+15}{2*2} =0,5; x_2=\frac{-13-15}{2*2} =-7$
x∈(-∞;-7)∪(0,5;+∞)
3. $2(1-x) \geq 5x(3x+2)$
$2-2x \geq 15 x^{2} +10x$
$2-2x-15 x^{2} -10x \geq 0$
$-15 x^{2} -12x+2 \geq 0$
$D=(-12)^2-4*(-15)*2=144+120=264$
$x_1= \frac{12+2 \sqrt{66} }{-30}= -\frac{6+ \sqrt{66} }{15} ; x_= \frac{12-2 \sqrt{66} }{-30}= -\frac{6- \sqrt{66} }{15}$
x∈ $[-\frac{6+ \sqrt{66} }{15}; -\frac{6- \sqrt{66} }{15} ]$
4.    $3 x^{2} +5x-8 \geq 0$
$D=25-4*3*(-8)=25+96=121$
$x_1= \frac{-5+11}{2*3} =1; x_2= \frac{-5-11}{2*3} =- \frac{8}{3}$
x∈(-∞;-8/3]∪[1;+∞)