Question

Решите уравнение:
а) $frac{x^{2}-3x-4}{x+1}=0;$
б) $x+7= frac{8}{x};$
в) $frac{x}{x+2}+ frac{x+2}{x-2}= frac{8}{ x^{2}-4}$

Answer 1

$1); frac{x^2-3x-4}{x+1}=0; ; Rightarrow ; ; left { {{x^2-3x-4=0} atop {x+1ne 0}} right. ; left { {x_1=-1,; x_2=4} atop {xne -1}} right. ; Rightarrow ; underline {x=4}\\2); x+7=frac{8}{x}; ; Rightarrow ; ; frac{x^2+7x-8}{x} =0; ,; left { {{x^2+7x-8=0} atop {xne 0}} right. ; ,; left { {{x_1=-8,; x_2=1} atop {xne 0}} right. ; Rightarrow \\underline {x_1=-8,; x_2=1}$

$3); frac{x}{x+2} + frac{x+2}{x-2} = frac{8}{x^2-4} ; ,; ; ODZ:; left { {{x+2ne 0,; x-2ne 0} atop {x^2-4ne 0}} right. ; Rightarrow ; ; xne pm 2\\ frac{x(x-2)+(x+2)^2-8}{(x+2)(x-2)} =0\\ frac{x^2-2x+x^2+4x+4-8}{(x+2)(x-2)} =0\\ frac{2x^2+2x-4}{(x+2)(x-2)} =0; ,; ; frac{2(x^2+x-2)}{(x+2)(x-2)} =0; ,; frac{2(x-1)(x+2)}{(x+2)(x-2)} =0; Rightarrow \\frac{2(x-1)}{x-2}=0; ; Rightarrow ; ; underline {x=1}$