Question

СРОЧНО,ПОМОГИТЕ РЕШИТЬ

Answer 1

$x^2+4x-5=0\\ D=16-4\cdot1\cdot(-5)=16+20=36\\ x_1= \frac{-4+6}{2} =1\\ x_2= \frac{-4-6}{2} =-5\\ x^2+4x-5=(x-1)(x+5)\\ ========================\\ x^2-3x+2=0\\ x_1+x_2=3\\ x_1\cdot x_2=2\\ x_1=1; x_2=2\\ x^2-3x+2=(x-1)(x-2)\\ ========================\\ 5x^2-4x-1=0\\ D=16-4\cdot5\cdot(-1)=36\\ x_1= \frac{4+6}{10} =1\\ x_2= \frac{4-6}{10} =- \frac{1}{5} \\ 5x^2-4x-1=5(x-1)(x+\frac{1}{5})\\========================$
$3x^2+2x-5=0\\ D=4-4\cdot3\cdot(-5)=64\\ x_1= \frac{-2+8}{6}=1\\ x_2= \frac{-2-8}{6}=- \frac{10}{6}=- \frac{5}{3}=-1 \frac{2}{3} \\ 3x^2+2x-5=3(x-1)(x+1 \frac{2}{3} )$