Question

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

Answer 1

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

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

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

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