\(1)\ \frac{x-2}{x+1}-\frac{5}{1-x}=\frac{x^2+27}{x^2-1};\)
\(\phantom{1)\ }\frac{x-2}{x+1}+\frac{5}{x-1}-\frac{x^2+27}{ (x-1)(x+1)}=0;\)
\(\phantom{1)\ }\frac{ (x-2)(x-1)+5(x+1)-(x^2+27)}{ (x-1)(x+1)}=0;\)
\(\phantom{1)\ }\frac{x^2-x-2x+2+5x+5-x^2-27}{ (x-1)(x+1)}=0;\)
\(\phantom{1)\ }\frac{\vphantom{^0}2x-20}{ (x-1)(x+1)}=0;\)
\(\phantom{1)\ }\begin{cases}2x-20=0,\\[-1ex]\\(x-1)(x+1)\ne0;\end{cases}\)
\(\phantom{1)\ }\begin{cases}x=10,\\[-1ex]\\x\ne1,\\[-1ex]\\x\ne{-}1.\end{cases}\)
Ответ: \(10.\)
\(2)\ \frac{3x+1}{3x-1}-\frac{3x-1}{3x+1}=\frac{6}{1-9x^2};\)
\(\phantom{2)\ }\frac{3x+1}{3x-1}-\frac{3x-1}{3x+1}-\frac{6}{1-9x^2}=0;\)
\(\phantom{2)\ }\frac{\vphantom{ ()}3x+1}{3x-1}-\frac{3x-1}{3x+1}+\frac{6}{9x^2-1}=0;\)
\(\phantom{2)\ }\frac{\vphantom{^0}3x+1}{3x-1}-\frac{3x-1}{3x+1}+\frac{6}{ (3x-1)(3x+1)}=0;\)
\(\phantom{2)\ }\frac{ (3x+1)^2-(3x-1)^2+6}{ (3x-1)(3x+1)}=0;\)
\(\phantom{2)\ }\frac{9x^2+6x+1-(9x^2-6x+1)+6}{ (3x-1)(3x+1)}=0;\)
\(\phantom{2)\ }\frac{9x^2+6x+1-9x^2+6x-1+6}{ (3x-1)(3x+1)}=0;\)
\(\phantom{2)\ }\frac{12x+6}{ (3x-1)(3x+1)}=0;\)
\(\phantom{2)\ }\begin{cases}12x+6=0,\\[-1ex]\\(3x-1)(3x+1)\ne0;\end{cases}\)
\(\phantom{2)\ }\begin{cases}x={-}\frac{1}{2},\\[-1ex]\\x\ne\frac{1}{3},\\[-1ex]\\x\ne{-}\frac{1}{3}.\end{cases}\)
Ответ: \({-}\frac{1}{2}.\)
\(3)\ \frac{\vphantom{^0}4}{x-3}+\frac{1}{x}=\frac{5}{x-2};\)
\(\phantom{3)\ }\frac{\vphantom{^0}4}{x-3}+\frac{1}{x}-\frac{5}{x-2}=0;\)
\(\phantom{3)\ }\frac{\vphantom{^0}4x(x-2)+(x-3)(x-2)-5x(x-3)}{ x(x-3)(x-2)}=0;\)
\(\phantom{3)\ }\frac{4x^2-8x+x^2-2x-3x+6-5x^2+15x}{ x(x-3)(x-2)}=0;\)
\(\phantom{3)\ }\frac{\vphantom{^0}2x+6}{ x(x-3)(x-2)}=0;\)
\(\phantom{3)\ }\begin{cases}2x+6=0,\\[-1ex]\\x(x-3)(x-2)\ne0;\end{cases}\)
\(\phantom{3)\ }\begin{cases}x={-}3,\\[-1ex]\\x\ne0,\\[-1ex]\\x\ne3,\\[-1ex]\\x\ne2.\end{cases}\)
Ответ: \({-}3.\)
\(4)\ \frac{2x^2-2x}{x^2-4}+\frac{6}{x+2}=\frac{x+2}{x-2};\)
\(\phantom{4)\ }\frac{2x^2-2x}{ (x-2)(x+2)}+\frac{6}{x+2}-\frac{x+2}{x-2}=0;\)
\(\phantom{4)\ }\frac{2x^2-2x+6(x-2)-(x+2)^2}{ (x-2)(x+2)}=0;\)
\(\phantom{4)\ }\frac{2x^2-2x+6x-12-(x^2+4x+4)}{ (x-2)(x+2)}=0;\)
\(\phantom{4)\ }\frac{2x^2-2x+6x-12-x^2-4x-4}{ (x-2)(x+2)}=0;\)
\(\phantom{4)\ }\frac{x^2-16}{ (x-2)(x+2)}=0;\)
\(\phantom{4)\ }\begin{cases}x^2-16=0,\\[-1ex]\\(x-2)(x+2)\ne0;\end{cases}\)
\(\phantom{4)\ }\begin{cases}x=4,\\[-1ex]\\x={-}4,\\[-1ex]\\x\ne2,\\[-1ex]\\x\ne{-2}.\end{cases}\)
Ответ: \(4,\ {-}4.\)
\(5)\ \frac{7}{x^2+2x}+\frac{x+1}{x^2-2x}=\frac{x+4}{x^2-4};\)
\(\phantom{5)\ }\frac{7}{ x(x+2)}+\frac{x+1}{ x(x-2)}-\frac{x+4}{ (x-2)(x+2)}=0;\)
\(\phantom{5)\ }\frac{ 7(x-2)+(x+1)(x+2)-x(x+4)}{ x(x-2)(x+2)}=0;\)
\(\phantom{5)\ }\frac{7x-14+x^2+2x+x+2-x^2-4x}{ x(x-2)(x+2)}=0;\)
\(\phantom{5)\ }\frac{\vphantom{^0}6x-12}{ x(x-2)(x+2)}=0;\)
\(\phantom{5)\ }\frac{\vphantom{^0} 6(x-2)}{ x(x-2)(x+2)}=0;\)
\(\phantom{5)\ }\frac{\vphantom{^0}6}{ x(x+2)}=0.\)
Ответ: корней нет.
\(6)\ \frac{x^2-9x+50}{x^2-5x}=\frac{x+1}{x-5}+\frac{x-5}{x};\)
\(\phantom{6)\ }\frac{x^2-9x+50}{ x(x-5)}-\frac{x+1}{x-5}-\frac{x-5}{x}=0;\)
\(\phantom{6)\ }\frac{x^2-9x+50-x(x+1)-(x-5)^2}{ x(x-5)}=0;\)
\(\phantom{6)\ }\frac{x^2-9x+50-x^2-x-(x^2-10x+25)}{ x(x-5)}=0;\)
\(\phantom{6)\ }\frac{x^2-9x+50-x^2-x-x^2+10x-25}{ x(x-5)}=0;\)
\(\phantom{6)\ }\frac{{-}x^2+25}{ x(x-5)}=0;\)
\(\phantom{6)\ }\frac{ {-}(x^2-25)}{ x(x-5)}=0;\)
\(\phantom{6)\ }\frac{ {-}(x-5)(x+5)}{ x(x-5)}=0;\)
\(\phantom{6)\ }\frac{ {-}(x+5)}{x}=0;\)
\(\phantom{6)\ }\frac{ {-}x-5}{x}=0;\)
\(\phantom{6)\ }\begin{cases}{-}x-5=0,\\[-1ex]\\x\ne0;\end{cases}\)
\(\phantom{6)\ }\begin{cases}x={-}5,\\[-1ex]\\x\ne0.\end{cases}\)
Ответ: \({-}5.\)
