Question

решите уравнение подробно пожалуйста
ответ:-13/3
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Answer 1

$\frac{(x+1)^{2}}{x-1} -\frac{20}{x^{2}+3x-4}+ \frac{x+1}{x+4}=x+1$

$x^{2} +3x-4=0\\x_{1}=-4,\\x_{2}=1.$

$\frac{(x+1)^{2}}{x-1} -\frac{20}{(x-1)*(x+4)}+ \frac{x+1}{x+4}=x+1,\\\\\frac{(x+1)^{2}*(x+4)}{(x-1)*(x+4)} -\frac{20}{(x-1)*(x+4)}+ \frac{(x+1)*(x-1)}{(x+4)*(x-1)}=x+1,\\\\\frac{(x+1)^{2}*(x+4)+(x+1)*(x-1)-20}{(x-1)*(x+4)}=x+1,\\\\(x+1)^{2}*(x+4)+(x+1)*(x-1)-20=(x+1)*(x-1)*(x+4),\\\\(x+1)^{2}*(x+4)+(x+1)*(x-1)-(x+1)*(x-1)*(x+4)=20,\\\\(x+1)*((x+4)*(x+1)-(x+4)*(x-1)+(x-1))=20,\\\\(x+1)*((x+4)*(x+1)-(x-1))+x-1)=20,\\\\(x+1)*((x+4)*(x-x+1+1)+x-1)=20,\\\\(x+1)*(2*(x+4)+x-1)=20,$

$(x+1)*(2x+8+x-1)=20,\\\\(x+1)*(3x+7)=20,\\\\3x^{2}+3x+7x+7-20=0,\\\\3x^{2}+10x-13=0.$

$D=b^{2}-4ac=10^{2}-4*3*(-13)=100+156=256 ;\\\\x_{1}= \frac{-b+\sqrt{D}}{2a}=\frac{-10+\sqrt{256}}{6}=\frac{-10+16}{6}=\frac{6}{6}=1;\\\\ x_{2}= \frac{-b-\sqrt{D}}{2a}=\frac{-10-\sqrt{256}}{6}=\frac{-10-16}{6}=\frac{-26}{6}=-\frac{13}{3}=-4\frac{1}{3} .$

Answer 2

Ответ:

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