Question

Решите уравнения пожалуйста! Даю 100 баллов! ​

Answer 1

$1)\frac{13}{(2x+7)(x-3)}+\frac{1}{2x+7} -\frac{6}{(x-3)(x+3)}=0\\\\\frac{13*(x+3)+(x-3)(x+3)-6*(2x+7)}{(2x+7)(x-3)(x+3)}=0\\\\\frac{13x+39+x^{2}-9-12x-42 }{(2x+7)(x-3)(x+3)}=0\\\\\frac{x^{2}+x-12 }{(2x+7)(x-3)(x+3)}=0\\\\2x+7\neq0 \ \Rightarrow \ x\neq-3,5\\\\x-3\neq0 \ \Rightarrow \ x\neq 3 \\\\x+3\neq0 \ \Rightarrow \ x\neq- 3 \\\\x^{2} +x-12=0\\\\D=1^{2}-4*(-12)=1+ 48=49=7^{2}\\\\x_{1} =\frac{-1-7}{2} =-4\\\\x_{2}=\frac{-1+7}{2} =3\\\\Otvet:\boxed{-4 \ ; \ 3}$

$2)\frac{x}{x-2} +\frac{x+2}{x+4} -\frac{12}{x^{2}+2x-8 }=\frac{x}{x-2} +\frac{x+2}{x+4} -\frac{12}{(x+4)(x-2)}=\frac{x*(x+4)+(x+2)*(x-2)-12}{(x+4)(x-2)} =\\\\=\frac{x^{2}+4x+x^{2}-4-12 }{(x+4)(x-2)}=\frac{2x^{2}+4x-16 }{x^{2}+2x-8 }=\frac{2(x^{2}+2x-8) }{x^{2}+2x-8 }=\boxed2$

$3)\frac{2}{x+3}+\frac{x^{2}+5 }{x^{3}+27 }-\frac{5x}{x^{2}-3x+9 } =\frac{2}{x+3}+\frac{x^{2}+5 }{(x+3)(x^{2}-3x+9) }-\frac{5x}{x^{2} -3x+9}=\\\\=\frac{2*(x^{2}-3x+9)+x^{2}+5-5x*(x+3) }{(x+3)(x^{2}-3x+9) }=\frac{2x^{2}-6x+18+x^{2}+5-5x^{2}-15x}{(x+3)(x^{2}-3x+9) } =\\\\=\frac{-2x^{2} -21x+23}{(x+3)(x^{2}-3x+9) }=-\frac{2x^{2}+21x-23 }{(x+3)(x^{2}-3x+9) } =-\frac{2x^{2}+21x-23 }{x^{3}+27 }$