Verified answer

[tex]\displaystyle 1)|x|(2x-7)=0\\|x|=0;2x-7=0\\\\x_1=0,x_2=\frac{7}{2}=3,5\\ \\2)(|x|-2,4)(3x+9,6)=0\\\\(|x|-\frac{12}{5})(3x+\frac{48}{5})=0\\ \\|x|-\frac{12}{5}=0;3x+\frac{48}{5}=0\\ \\x_1=-\frac{16}{5}=-3,2;x_2=-\frac{12}{5}=-2,4;x_3=\frac{12}{5}=2,4[/tex]