Решите неравенство двумя способами 6x²-7x+2>0 8x²+10x-3≥0 49x²-28x+4<0 4x²-4x+15≤0

Question 1

Question 2

Решите задачу:

$I. \ \ 6x^2-7x+2\ \textgreater \ 0, \\ a=6\ \textgreater \ 0, \\ 6x^2-7x+2=0, \\ D=1\ \textgreater \ 0, \\ x_1=\frac{1}{2}, \ x_2=\frac{2}{3}, \\ \left [ {{x\ \textless \ \frac{1}{2},} \atop {x\ \textgreater \ \frac{2}{3};}} \right. \\ x\in(-\infty;\frac{1}{2})\cup(\frac{2}{3};+\infty); \\ \\ II. \ \ 6x^2-7x+2\ \textgreater \ 0, \\ 6x^2-7x+2=0, \\ D=1, \\ x_1=\frac{1}{2}, \ x_2=\frac{2}{3}, \\ 6(x-\frac{1}{2})(x-\frac{2}{3})\ \textgreater \ 0, \\ (x-\frac{1}{2})(x-\frac{2}{3})\ \textgreater \ 0, \\ x\in(-\infty;\frac{1}{2})\cup(\frac{2}{3};+\infty);$

$I. \ 8x^2+10x-3 \geq 0, \\ a=8\ \textgreater \ 0, \\ 8x^2+10x-3=0, \\ D_1=1\ \textgreater \ 0, \\ x_1=-\frac{3}{4}, \ x_2=-\frac{1}{2}, \\ \left [ {{x \leq -\frac{3}{4},} \atop {x \geq -\frac{1}{2},}} \right. \\ x\in(-\infty;-\frac{3}{4})\cup(-\frac{1}{2};+\infty); \\ \\ II. \ 8x^2+10x-3 \geq 0, \\ 8x^2+10x-3=0, \\ D_1=1\ \textgreater \ 0, \\ x_1=-\frac{3}{4}, \ x_2=-\frac{1}{2}, \\ 8(x+\frac{3}{4})(x+\frac{3}{4}) \geq 0, \\ (x+\frac{3}{4})(x+\frac{3}{4}) \geq 0, \\ x\in(-\infty;-\frac{3}{4})\cup(-\frac{1}{2};+\infty);$

$I. \ 49x^2-28x+4\ \textless \ 0, \\ a=49\ \textgreater \ 0, \\ 49x^2-28x+4=0, \\ D_1=0, \\ x_1=x_2=\frac{14}{49} \\ x\in\varnothing; \\ \\ II. \ 49x^2-28x+4\ \textless \ 0, \\ (7x-2)^2\ \textless \ 0, \\ x\in\varnothing;$

$I. \ 4x^2-4x+15 \leq 0, \\ a=4\ \textgreater \ 0, \\ D_1=-56\ \textless \ 0, \\ x\in\varnothing; \\ \\ II. \ 4x^2-4x+15 \leq 0, \\ (2x+1)^2+14 \leq 0, \\ (2x+1)^2 \leq -14, \\ x\in\varnothing.$