Помогите пожалуйста номер 184

Question

Алгебра

Юлия

7 год назад

Помогите пожалуйста номер 184

ОТВЕТЫ

Владислав

Jul 5, 2019

$\frac{30}{ x^{2} -1} - \frac{7+18x}{x^3-1} = \frac{13}{ x^{2} +x+1} \\ \frac{30}{(x-1)(x+1)} - \frac{7+18x}{(x-1)( x^{2} +x+1)} - \frac{13}{ x^{2} +x+1} =0 \\ \frac{30( x^{2} +x+1)-(7+18x)(x+1)-13( x^{2} -1)}{(x-1)^2(x+1)( x^{2} +x+1)} =0 \\ \frac{30 x^{2} +30x+30-7x-7-18 x^{2} -18x-13 x^{2} +13}{(x-1)^2(x+1)( x^{2} +x+1)} =0 \\ \frac{- x^{2}+5x +36}{(x+1)(x-1)^2( x^{2} +x+1)}=0$
$\left \{ {{x^2-5x-36=0} \atop {(x-1)^2(x+1)( x^{2} +x+1) \neq0 }} \right. \left \{ {{x=-4,x=9} \atop {x \neq 1,x \neq -1}} \right. \\ OTBET:-4,9$

$\frac{2 x^{2} +7}{ x^{3}+1 } - \frac{2}{x^2-x+1}= \frac{3}{x+1} \\ \frac{2x^2+7}{(x+1)(x^2-x+1)} - \frac{2}{x^2-x+1}- \frac{3}{x+1} =0 \\ \frac{2x^2+7-2(x+1)-3(x^2-x+1)}{(x+1)(x^2-x+1)} =0 \\ \frac{2x^2+7-2x-2-3x^2+3x-3}{x^3+1} =0 \\ \frac{-x^2+x+2}{x^3+1} =0 \\ \left \{ {{x^2-x-2=0} \atop {x^3+1 \neq 0}} \right. \\ \left \{ {{x=2,x=-1} \atop {x \neq -1}} \right. \\ OTBET: 2$

$\frac{2y-1}{14y^2+7y} - \frac{2y+1}{6y^2-3y} = \frac{8}{3-12y^2} \\ \frac{2y-1}{7y(2y+1)} - \frac{3y+1}{3y(2y-1)} - \frac{8}{3(1-2y)(1+2y)}=0 \\ \frac{2y-1}{7y(2y+1)} - \frac{2y+1}{3y(2y-1)} + \frac{8}{3(2y-1)(2y+1)}=0 \\ \frac{3(2y-1)(2y-1)-7(2y+1)(2y+1)+8*7y}{21y(2y-1)(2y+1)} =0 \\ \frac{12y^2-12y+3-28y^2-28y-7+56y}{21y(2y-1)(2y+1)}=0 \\ \frac{-16y^2+16y-4}{21y(2y-1)(2y+1)} =0$
$\left \{ {{16y^2-16y+4=0} \atop {21y(2y-1)(2y+1) \neq 0}} \right. \\ \left \{ {{16y^2-16y+4=0} \atop {21y(2y-1)(2y+1) \neq 0}} \right. \\ \left \{ {{4y^2-4y+1=0} \atop { 21y(2y-1)(2y+1) \neq 0}} \right. } \right. \\ \left \{ {{y= \frac{1}{2} } \atop {y \neq 0,y \neq \frac{1}{2} ,y \neq - \frac{1}{2} }} \right.$
OTBET : нет корней.

$\frac{1}{x+1} - \frac{1}{x^2-3x+2} = \frac{6}{2-x-2x^2+x^3} \\ \frac{1}{x+1} - \frac{1}{(x-1)(x-2)} - \frac{6}{(x-2)(x-1)(x+1)} =0 \\ \frac{(x-2)(x-1)-(x+1)-6}{(x-2)(x-1)(x+1)} =0 \\ \frac{x^2-3x+2-x-1-6}{(x-2)(x-1)(x+1)} =0 \\ \frac{x^2-4x-5}{(x-2)(x-1)(x+1)} =0 \\ \left \{ {{x^2-4x-5=0} \atop {(x-2)(x-1)(x+1) \neq 0}} \right. \\ \left \{ {{x=-1,x=5} \atop {x \neq 2,x \neq 1,x \neq -1}} \right. \\ OTBET:5$

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