Question

помогите пожалуйста

Answer 1

$(x^2-9)^2-4(x^2-9)+3=0\(x^2-9)=y\y^2-4y+3=0\ left { {y_2y_2=3} atop {y_1+y_2=4}} right. [ { {{y=1} atop {y=3}} right.\x^2-9=1\x^2=10 [{ {{x= sqrt{10} } atop {x=- sqrt{10} }} right. \\x^2-9=3\x^2=12[{ {{x= sqrt{12}=2 sqrt{3} } atop {x=- sqrt{12}=-2 sqrt{3} }} right.$
Ответ: $-2 sqrt{3};- sqrt{10}; sqrt{10};2 sqrt{3}$

$(x^2+x+1)(x^2+x+2)=12\(x^2+x+1)=y\y(y+1)=12\y^2+y-12=0\ left { {{y_1y_2=-12} atop {y_1+y_2=-1}} right. [{ {{y=-4} atop {y=3}} right. \x^2+x+1=-4\x^2+x+5=0\\x^2+x+1=3\x^2+x-2=0\ left { {{x_1x_2=-2} atop {x_1+x_2=-1}} right. [{ {{x=-2} atop {x=1}} right.$
Ответ: $-2;1$

$16x^2-25x^2+9=0\-9x^2=-9\x^2=1\ { {{x=-1} atop {x=1}} right.$
Ответ: $-1;1$

$x^4+5x^2-36=0\x^2=y\y^2+5y-36=0\ left { {{y_1y_2=-36} atop {y_1+y_2=-5}} right. [{ {{y=-9} atop {y=4}} right. \x^2=-9\\x^2=4\ { {{x=2} atop {x=-2}} right.$
Ответ: $-2;2$

Answer 2

4 пример я не доделал