Question

Решите уравнения:

а) $\frac{3y+8}{7} - \frac{y-6}{2} = 2$

б) $\frac{2x^{2}+3x+1 }{x^{3}+2x^{2} +3x+2 } =0$

Answer 1

Відповідь:

a)  y=30

b)$x=-\frac{1}{2}$

Пояснення:

a)

$\frac{3y+8}{7} -\frac{y-6}{2} =2\\\\2(3y+8)-7(y-6)=28\\\\\\6y+16-7y+42=28\\\\-y+58=28\\\\-y=28-58\\\\-y=-30\\\\y=30$

b)

$\frac{2x^2+3x+1}{x^3+2x^2+3x+2} =0, x\neq -1\\\\ \frac{2x^2+2x+x+1}{x^3+x^2+x^2+x+2x+2} =0\\\\\frac{2x(x+1)+x+1}{x^2(x+1)+x(x+1)+2(x+1)} =0\\\\\frac{(x+1)(2x+1)}{(x+1)(x^2+x+2)} =0\\\\\frac{2x+1}{x^2+x+2} =0\\\\2x+1=0\\\\2x=-1\\\\x=-\frac{1}{2} , x\neq -1$