Question

помогите плиз

Answer 1

$\frac{6}{x} + \frac{6}{x+1}-5=0 \\ \frac{6(x+1)+6x-5x(x+1)}{x(x+1)} =0 \\ \frac{6x+6+6x-5x^2-5x}{x(x+1)} =0 \\ \frac{-5x^2+7x+6}{x(x+1)} =0 \\ \left \{ {{5x^2-7x-6=0} \atop {x \neq 0; -1}} \right. \\ D=b^2-4ac=49+120=169=13^2 \\ x_{1,2}= \frac{-b\pm\sqrt{D}}{2a} \\ x_{1}= \frac{7+13}{10}=2 \\ x_{2}= \frac{7-13}{10} =-0.6 \\ \left \{ {{x=-0.6;2} \atop {x \neq -1;0}} \right. \\ x=-0.6;2$

Ответ: $x=-0.6; 2$

Answer 2

$\frac{6(x+1)+6x}{x(x+1)} =5$
$\frac{6x+6+6x}{ x^{2} +x} =5$
$12x+6=5 x^{2} +5x$
$5 x^{2} -7x-6=0$
$D= 7^{2} -4*5*(-6)=49+120=169$
$x_{1} = \frac{7-13}{10}=- \frac{6}{10}=-0.6$
$x_{2} = \frac{7+13}{10} =2$