Question

Помогите очень надо !!!!

Answer 1

$\displaystyle 1)10x^2+3x=1\\10x^2+3x-1=0\\10x^2+5x-2x-1=0\\5x(2x+1)-(2x+1)=0\\(2x+1)(5x-1)=0\\2x+1=0,;5x-1=0\\\\x_1=-\frac{1}{2},x_2=\frac{1}{5}\\ \\2)3x^2-1=2x\\3x^2-1-2x=0\\3x^2-2x-1=0\\3x^2+x-3x-1=0\\x(3x+1)-(3x+1)=0\\(3x+1)(x-1)=0\\3x+1=0,x-1=0\\\\x_1=-\frac{1}{3},x_2=1\\ \\3)25x^2+60x+36=0\\(5x+6)^2=0\\5x+6=0\\5x=-6\\\\x=-\frac{6}{5}=-1\frac{1}{5}=-1,2$

$\displaystyle 4)3x^2-3x+14=0\\\\x=\frac{-(-3)б\sqrt{(-3)^2-4*3*14} }{2*3}=\frac{3б\sqrt{-159} }{6}\\ \\x \notin R$