c)
[tex]\frac{x^2-5x+4}{5}=\frac{1-3x}{2}\ |*10\\2(x^2-5x+4)=5(1-3x)\\2x^2-10x+8=5-15x\\2x^2-10x+8-5+15x=0\\2x^2+5x+3=0\\\Delta=5^2-4*2*3=25-24=1+\sqrt\Delta=1\\x_1=\frac{-5-1}{2*2}=\frac{-6}{4}=-\frac{3}{2}\\x_2=\frac{-5+1}{2*2}=\frac{-4}{4}=-1\\x\in\{-\frac{3}{2},-1\}[/tex]
d)
[tex]\frac{(x+1)^2}{3}=\frac{(x+4)^2}{7}-1\\\frac{x^2+2x+1}{3}=\frac{x^2+8x+16}{7}-1\ |*21\\7(x^2+2x+1)=3(x^2+8x+16)-21\\7x^2+14x+7=3x^2+24x+48-21\\7x^2+14x+7-3x^2-24x-48+21=0\\4x^2-10x-20=0\ |:2\\2x^2-5x-10=0\\\Delta=(-5)^2-4*2*(-10)=25+80=105\\\sqrt\Delta=\sqrt{105}\\x_1=\frac{5-\sqrt{105}}{2*2}=\frac{5-\sqrt{105}}{4}\\x_2=\frac{5+\sqrt{105}}{2*2}=\frac{5+\sqrt{105}}{4}\\x\in\{\frac{5-\sqrt{105}}{4},\frac{5+\sqrt{105}}{4}\}[/tex]
