0;\\\\x^2+a-1=0;\\\\x^2=1-a;\\\\1-a \geq 0; a <1; x_1=-\sqrt{1-a};x_2=\sqrt{1-a};\\\\a=1;x_1=x_2=0;" alt="x^4+ax^2+a-1=0;\\\\x^4+(a-1)x^2+x^2+a-1=0;\\\\x^2(x^2+a-1)+1*(x^2+a-1)=0;\\\\(x^2+1)(x^2+a-1)=0;\\\\x^2+1>0;\\\\x^2+a-1=0;\\\\x^2=1-a;\\\\1-a \geq 0; a <1; x_1=-\sqrt{1-a};x_2=\sqrt{1-a};\\\\a=1;x_1=x_2=0;" align="absmiddle" class="latex-formula">
ответ: при a<1</p>