Помогите решить 1б, пожалуйста. Срочно.

Question 1

Помогите решить 1б, пожалуйста.
Срочно.

Question 2

$x^{3}-25x=0; x( x^{2} -25)=0 \\ \left \{ {{x=0} \atop { x^{2}-25 =0}} \right. \left \{ {{x=0} \atop { x^{2} =25}} \right. \left \{ {{x=0} \atop {x=+-5}} \right.$
$\frac{3y+2}{4 y^{2}+y } + \frac{y-3}{16 y^{2}-1 }= \frac{3}{4y-1} \\ \\ \frac{3y+2}{y(4y+1)} + \frac{y-3}{(4y-1)(4y+1)} - \frac{3}{4y-1}=0 \\ \\ \frac{(3y+2)(4y-1)+y(y-3)-3y(4y+1)}{y(4y-1)(4y+1)} =0 \\ \\ \frac{ y^{2-y-2} }{y(4y-1)(4y+1)} =0 \\ \\ y^{2}-y-2=0, y \neq 0,y \neq +- \frac{1}{4} \\ \\ D=1-4*1*(-2)=9 \\ \\ x_{1} = \frac{1-3}{2}=-1; x_{2}= \frac{1+3}{2}=2$
2.
$2 x^{2} -x-15 \ \textgreater \ 0 \\ \\ D = 1 - 4*2*(-15)=121 \\ \\ x_{1} = \frac{1-11}{4}=-2,5 ;x_{2}= \frac{1+11}{4}=3$
x∈(-∞;2.5)∪(3;∞)
$x^{2} \ \textless \ 16 \\ x=+-4$
x∈(-∞;-4)∪(4;∞)
3.
$(x+11)(x+2)(x-9)\ \textless \ 0$
x∈(-∞;-11)∪(-2;9)
$\frac{x+3}{x-8} \ \textgreater \ 0$
x∈(-∞;-3)∪(8;∞)
4.
$x^{4} -4 x^{2} -45=0 \\ \\ t= x^{2} \\ \\ t^{2} -4t -45 = 0\\ \\ D = 16 -4 *1(-45) = 196 \\ \\ t_{1} = \frac{4-14}{2}= 5; t_{2}= \frac{4+14}{2}=9 \\ \\ \left \{ {{ x^{2} =5} \atop { x^{2} =9}} \right. \left \{ {{x= \sqrt{5} } \atop {x=+-3}} \right.$
5.
$2 x^{2} +nx + 8=0 \\ \\ n^{2}-4*2*8\ \textless \ 0 \\ \\ n^{2}- 64\ \textless \ 0 \\ \\ n^{2}\ \textless \ 64 \\ \\ n\ \textless \ 8$
n∈(-∞;8)
6.
$y= \sqrt{3x-2 x^{2} } \\ \\ |3x-2 x^{2} | \geq 0 \\ \\ \left \{ {{3x-2 x^{2} \geq 0} \atop {2 x^{2} -3x \geq 0}} \right. \\ \\ 3x-2 x^{2} \geq 0; x(3-2x) \geq 0 \\ \\ \left \{ {{x \geq 0} \atop {3-2x \geq 0}} \right. \left \{ {{x \geq =0} \atop {x \leq 1.5}} \right. \\ \\ 2x^{2}-3x\geq 0 ;x(2x-3) \geq 0\\ \\ \left \{ {{x \geq 0} \atop {x \geq 1.5}} \right.$
x∈[-1.5;1.5]