Question

Помогите пожалуйста, я эту тему не понял просто​

Answer 1

$1.a)\;x_1=2\frac{1}{8}=2,125 ;\;x_2=-0,2\\\\(x-x_1)(x-x_2)=0\\(x-2,125)(x+0,2)=0\\x^2-2,125x+0,2x-0,425=0\\x^2-1,925x-0,425=0\\\\1.b)\;x_1=3-\sqrt{5};\;x_2=3+\sqrt{5}\\\\(x-x_1)(x-x_2)=0\\(x-(3-\sqrt{5}))(x-(3+\sqrt{5})=0\\x^2-(3-\sqrt{5})x-(3+\sqrt{5})x+(3-\sqrt{5})(3+\sqrt{5})=0\\x^2-3x+\sqrt{5}x-3x-\sqrt{5}x+9-5=0\\x^2-6x+4=0$

$1.c)\;x_1=x_2=-1\frac{3}{5}=-1,6\\\\(x-x_1)(x-x_2)=0\\(x+1,6)(x+1,6)=0\\(x+1,6)^2=0\\x^2+3,2x+2,56=0$

$2.\;3x^2-x-8=0|:3\\\\x^2-\frac{1}{3}x-\frac{8}{3}=0\\\\x_1+x_2=\frac{1}{3},\; \; \; \; x_1x_2=-8\\\\a)\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}=\frac{1/3}{-8}=-\frac{1}{24}\\\\b)x_1^2+x_2^2=(x_1^2+2x_1x_2+x_2^2)-2x_1x_2=(x_1+x_2)^2-2x_1x_2=\\=(1/3)^2-2*(-8)=1/9+16=16\frac{1}{9}\\\\c)\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{x_1^2+x_2^2}{x_1x_2}=\frac{16\frac{1}{9} }{-8}=\frac{\frac{145}{9} }{-8}=-\frac{145}{72}=-2\frac{1}{72}$

$d) y_1=\frac{1}{x_1};\;y_2=\frac{1}{x_2}\\\\(y-\frac{1}{x_1})(y-\frac{1}{x_2})=0\\\\y^2-(\frac{1}{x_1}+\frac{1}{x_2})y+\frac{1}{x_1x_2}=0$

$y^2-(-\frac{1}{24})y-\frac{1}{8}=0\\\\y^2+\frac{1}{24}y-\frac{1}{8}=0|*24\\\\24y^2+y-3=0$